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5-3 Solving Trigonometric Equations the interval , the only solution is when x = 1.34. 39. x log x +...

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Solve each equation for all values of x. 1. 5 sin x + 2 = sin x SOLUTION: The period of sine is 2π, so you only need to find solutions on the interval . The solutions on this interval are and . Solutions on the interval (, ), are found by adding integer multiples of 2π. Therefore, the general form of the solutions is +2nπ, + 2nπ, . 2. 5 = sec 2 x + 3 SOLUTION: The period of secant is 2π, so you only need to find solutions on the interval . The solutions on this interval are , , , and . Solutions on the interval (, ), are found by adding integer multiples of 2π. Therefore, the general form of the solutions is + 2nπ, + 2nπ, + 2nπ, + 2nπ, . 2 3. 2 = 4 cos 2 x + 1 SOLUTION: The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on this interval are , , , and . Solutions on the interval (, ), are found by adding integer multiples of 2π. Therefore, the general form of the solutions is + 2nπ, + 2nπ, + 2nπ, + 2nπ, . 4. 4 tan x 7 = 3 tan x 6 SOLUTION: The period of tangent is π, so you only need to find solutions on the interval . The only solution on this interval is . Solutions on the interval (, ), are found by adding integer multiples of π. Therefore, the general form of the solutions is + nπ, . 2 eSolutions Manual - Powered by Cognero Page 1 5-3 Solving Trigonometric Equations
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Page 1: 5-3 Solving Trigonometric Equations the interval , the only solution is when x = 1.34. 39. x log x + 5x cos x = –2 SOLUTION: On the interval , the solutions are when x = 1.84 and

Solve each equation for all values of x.1. 5 sin x + 2 = sin x

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is  + 2nπ,  + 2nπ, .

2. 5 =  sec 2 x + 3

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

3. 2 = 4 cos2 x + 1

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on

the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

4. 4 tan x – 7 = 3 tan x – 6

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

5. 9 + cot2 x = 12

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

6. 2 – 10 sec x = 4 – 9 sec x

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

7. 3 csc x = 2 csc x +

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

8. 11 =  3 csc2 x + 7

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are , , , and . Solutions

on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

9. 6 tan2 x – 2 = 4

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

10. 9 + sin2 x = 10

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the interval

(– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the solutions is

 + 2nπ,  + 2nπ, .

11. 7 cot x –   = 4 cot x

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

12. 7 cos x = 5 cos x +

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ,   + 2nπ, .

Find all solutions of each equation on [0, 2 ).13. sin4 x + 2 sin2 x − 3 = 0

SOLUTION:  

On the interval [0, 2π),  when x =  and 

 when x = . Since  is not a real 

number, the equation  yields no additional solutions.

14. –2 sin x = –sin x cos x

SOLUTION:  

The equation cos x = 2 has no real solutions since the maximum value the cosine function can obtain is 1. On the interval [0, 2π), the equation  has solutions 0 and π.

15. 4 cot x = cot x sin2 x

SOLUTION:  

The equations sin x = 2 and sin x = –2 have no real solutions. On the interval [0, 2π), the equation cot x =

0 has solutions  and  .

16. csc2 x – csc x + 9 = 11

SOLUTION:  

On the interval [0, 2π), the equation csc x = –1 has a

solution of  and the equation csc x = 2 has

solutions of    and  .

17. cos3 x + cos2 x – cos x = 1

SOLUTION:  

On the interval [0, 2π), the equation cos x = 1 has a solution of 0 and the equation cos x = –1 has a solution of  π.

18. 2 sin2 x = sin x + 1

SOLUTION:  

On the interval [0, 2π), the equation sin x = 1 has a

solution of  and the equation sin x =  has  

solutions of    and  .

19. TENNIS A tennis ball leaves a racquet and heads toward a net 40 feet away. The height of the net is the same height as the initial height of the tennis ball. 

a. If the ball is hit at 50 feet per second, neglecting air resistance, use  to find the interval of possible angles of the ball needed to clear the net. b. Find  if the initial velocity remained the same but the distance to the net was 50 feet.

SOLUTION:  a.

The interval is [15.4°, 74.6°].   b.

If the distance to the net is 50 feet, then the angle would be 19.9° or 70.1°.

20. SKIING In the Olympics aerial skiing competition, skiers speed down a slope that launches them into the air, as shown. The maximum height a skier can

reach is given by hpeak =  , where g is

the acceleration due to gravity or 9.8 meters per second squared.

a. If a skier obtains a height of 5 meters above the end of the ramp, what was the skier’s initial speed? b. Use your answer from part a to determine how long it took the skier to reach the maximum height if tpeak = .

SOLUTION:  a.

The skier’s initial speed was 11.67 meters per second.   b.

  The time it took the skier to reach the maximum height was about 1.0 seconds.

Find all solutions of each equation on the interval [0, 2 ).

21. 1 = cot 2 x + csc x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

22. sec x = tan x + 1

SOLUTION:  

   

  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is 0.

23. tan 2 x = 1 – sec x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are 0,

, and .

24. csc x + cot x = 1

SOLUTION:  

 

x =  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is .

25. 2 – 2 cos2 x = sin x + 1

SOLUTION:  

Check

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

26. cos x – 4 = sin x – 4

SOLUTION:   

 

Therefore, on the interval [0, 2π) the only valid

solutions are and .

27. 3 sin x = 3 – 3 cos x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the only valid

solutions are and 0.

28. cot2 x csc2 x – cot2 x = 9

SOLUTION:  

is undefined. The solutions of  are 

  Check

 

 

 

On the interval , there are no real values of x for which , but  for 

29. sec2 x – 1 + tan x –  tan x =

SOLUTION:  

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

30. sec2 x tan2 x + 3 sec2 x – 2 tan2 x = 3

SOLUTION:  

 

The square of any real number must be greater than or equal to zero so sec2 x = –1 has no solutions. Therefore, on the interval [0, 2π) the only solutions are 0 and π.

31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive power PR of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical components, PH and PV.

Using the equations above, determine for what values of  PV and PH are equivalent.

SOLUTION:  

The sine and cosine have the same values in the

interval [0, 2π) at  and  .

Therefore, the components will be equivalent when

.

Find all solutions of each equation on the interval [0, 2 ).

32.  + cos x = 2

SOLUTION:  

On the interval [0, 2π), cos x =  when x =  and 

when x = .

33.  +  = –4

SOLUTION:  

On the interval [0, 2π),cos x = – when x =

 and when x = .

34.  +  = cos x

SOLUTION:  

On the interval [0, 2π) there are no values of x for

which sin x = –2 and the sin x = 1 only when x = .

However, tan  is undefined, so there are no 

solutions for the original equation.

35. cot x cos x + 1 =  + 

SOLUTION:  

On ,  when x =  and when x =

.

GRAPHING CALCULATOR Solve each equation on the interval [0, 2 ) by graphing. Round to the nearest hundredth.

36. 3 cos 2x = ex + 1

SOLUTION:  

On the interval , the only solution is when x = 0.33.

37. sin x + cos x = 3x

SOLUTION:  

On the interval , the only solution is when x = 0.41.

38. x2 = 2 cos x + x

SOLUTION:  

On the interval , the only solution is when x = 1.34.

39. x log x + 5x cos x = –2

SOLUTION:  

On the interval , the solutions are when x = 1.84 and when x = 4.49.

40. METEOROLOGY The average daily temperature in degrees Fahrenheit for a city can be modeled by t

= 8.05 cos + 66.95, where x is a function of

time, x = 1 represents January 15, x = 2 represents February 15, and so on. a. Use a graphing calculator to estimate the temperature on January 31. b. Approximate the number of months that the average daily temperature is greater than 70 throughout the entire month. c. Estimate the highest temperature of the year and the month in which it occurs.

SOLUTION:  a. Use a graphing calculator to graph t = 8.05

cos + 66.95. Because x = 1 represents

January 15, x = 1.5 represents January 31. Use the TRACE feature to determine the temperature whenx = 1.5.

Therefore, the temperature on January 31 is about 61.3º.   b. Graph y = 70. Then use the intersect feature under the CALC menu to determine the interval on

which 8.05 cos + 66.95 > 70.

So, 8.05 cos + 66.95 > 70 on the interval

3.74 < x < 8.26. Because x = 3.74 corresponds to about the middle of April, the temperature does not exceed 70º for the entire month until May. Therefore, the average daily temperature is greater than 70º for the entire month during the months of May, June, and July. Therefore, the average temperature was higher than 70º for three months.   c. Use the maximum feature under the CALC menu to find the maximum value of the function on [0, 12].

The maximum value is 75 when x ≈ 6. Therefore, the maximum average monthly temperature is 75º during the month of June.

Find the x-intercepts of each graph on the interval [0, 2 ).

41. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) cos x = 0 when x =  and x

= .

42. 

 

SOLUTION:  Let y = 0 and solve for x.

  On the interval [0, 2π) sin x = 0 when x = 0 and x = π.

43. 

SOLUTION:  Let y = 0 and solve for x.

 

On the interval [0, 2π) cot x = 1 when x =  and x

= .

44. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) tan x = 0 when x = 0 and x =

π, tan x = –1 when x =  and x = , and tan x

= 1 when x =  and x = .

Find all solutions of each equation on the interval [0, 4 ).

45. 4 tan x = 2 sec2 x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

46. 2 sin2 x + 1 = –3 sin x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, , , and .

47. csc x cot2 x = csc x

SOLUTION:   

On the interval [0, 4π) the solutions are , ,

, , , , , and .

48. sec x + 5 = 2 sec x + 3

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

49. GEOMETRY Consider the circle below.

a. The length s of  is given by s = r(2 ) where 0≤   ≤  . When s = 18 and AB = 14, the radius is r

= . Use a graphing calculator to find the

measure of 2 in radians. b. The area of the shaded region is given by A =

. Use a graphing calculator to find the

radian measure of θ  if the radius is 5 inches and the area is 36 square inches. Round to the nearest hundredth.

SOLUTION:  a. Rewrite the arclength formula using s = 18 and r

= .

On the graphing calculator, find the intersection of Y1 = 18sinθ and Y2 = 14θ.

The value of 2θ = 2(1.1968) or about 2.39 radians.     b. First, substitute into the given area formula and rearrange it.

  Using a graphing calculator, find the intersection of Y1 = 2.88 and Y2 = θ – sin θ.

When the area is 36 square inches and the radius is 5 inches, then the measure of θ is 3.01 radians.

Solve each inequality on the interval [0, 2 ).50. 1 > 2 sin x

SOLUTION:  Graph y = 1 and y = 2 sin x on [0, 2π). Use the intersect feature under the CALC menu to determine on what interval(s) 1 > 2 sin x.

  The graphs intersect at about 0.524 or  and about 

2.618 or . The graph of y = 1 is above the other

graph when x < and when x > . Therefore, 1

> 2 sin x on 0 ≤ x <    or   < x < 2π.

51. 0 < 2 cos x –

SOLUTION:  Graph y = 2 cos x –  on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0 < 2 cos x – .

  The zeros of y = 2 cos x –  are about 0.785 or  and about 5.498 or  . The graph of y = 0 is

below the other graph when 0 ≤ x <  or   < x <

2π. Therefore, 0 < 2 cos x –  on 0 ≤ x <  or   < 

x < 2π.

52.  ≥ 

SOLUTION:  

Graph y = and y =  on [0, 2π). Use

the intersect feature under the CALC menu to

determine on what interval(s) ≥  .

  The graphs intersect at about 1.047 or  and about 

2.094 or . The graph of y = is below the

other graph when   ≤ x ≤  .

Therefore, ≥   on  ≤ x ≤  .

53.  ≤ tan x cot x

SOLUTION:  

Graph y = and y = tan x cot x on [0, 2π).

Use the intersect feature under the CALC menu to

determine on what interval(s) ≤ tan x cot

x.  

  The graphs intersect at about 3.142 or π. The graph of y = tan x cot x is at or below the other graph when 0 ≤ x < 2π.  

Therefore, ≤ tan x cot x on 0 ≤ x < 2π.

54. cos x ≤

SOLUTION:  

Graph y = cos x and y =  on [0, 2π). Use the

intersect feature under the CALC menu to

determine on what interval(s) cos x ≤ .

  The graphs intersect at about 2.618 or  and about

3.665 or . The graph of y = cos x is below the

other graph when  ≤ x ≤  .

 

Therefore, cos x ≤ on  ≤ x ≤  .

55.  sin x − 1 < 0

SOLUTION:  Graph y =  sin x − 1. Use the zero feature underthe CALC menu to determine on what interval(s)

 sin x − 1 < 0.

  The zeros of  sin x − 1 are about 0.785 or  and

about 2.356 or . The graph is below the x-axis

when 0 ≤ x <  or   < x < 2π.

Therefore,  sin x − 1 < 0 on 0 ≤ x <  or   < 

x < 2π.

56. REFRACTION When light travels from one medium to another it bends or refracts, as shown.  

Refraction is described by n2 sinθ1 = n1 sinθ2, wheren1 is the index of refraction of the medium the light

is entering, n2 is the index of refraction of the medium the light is exiting, θ1 is the angle of

incidence, and θ2 is the angle of refraction. a. Find θ2 for each material shown if the angle of incidence is 40º and the index of refraction for air is 1.00.

b. If the angle of incidence is doubled to 80 , will the resulting angles of refraction be twice as large asthose found in part a?

SOLUTION:  a. glass:

  ice:

  plastic:

  water:

  b. Find the angle of refraction for one of the mediums using an angle of incidence of 80 . glass:

θ2 ≈ 25 when θ1 = 40 andθ2 ≈ 40.4 when θ1 = 80 . Therefore, the angle of refraction is not twice as large.

57. ERROR ANALYSIS Vijay and Alicia are solving tan2 x – tan x +  =   tan x. Vijay thinks that

the solutions are x =  + n , x =  + n , x =

 + n , and x =  + n . Alicia thinks that the

solutions are x =  + n and x =  + n . Is

either of them correct? Explain your reasoning.

SOLUTION:  First, solve tan2 x – tan x +  =   tan x.

  On [0, 2π) tan x = 1 when x =  and x =  and 

tan x = when x =  and x = .

  Sample answer: Therefore, Vijay’s solutions are correct; however, they are not stated in the simplest

form. For example, his solutions of x =  + n and

x =  + n could simply be stated as x =  + 

n because when n = 1,  + nπ  is equivalent to

.

CHALLENGE  Solve each equation on the interval [0, 2π].

58. 16 sin5 x + 2 sin x = 12 sin3 x

SOLUTION:  

On [0, 2π) sin x = 0 when x = 0 and x = π, sin x =

when x =  and x = , sin x = when

x =  and x = , sin x =  – when x =  and 

x = , and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are 0, , , , , , , ,

, and .

59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3

SOLUTION:  

On [0, 2 ) sin x = 1 when x = , sin x  = –1 when

x = , sin x =  – when x =  and x = ,

and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are , , , , , and .

60. REASONING Are the solutions of csc x =  andcot2 x + 1 = 2 equivalent? If so, verify your answer algebraically. If not, explain your reasoning.

SOLUTION:  

 

 

The solutions to csc x =  are   and  . Using

a Pythagorean identity, cot2 x + 1 = 2 simplifies to csc2 x = 2 or csc x = ± . Therefore, there are

two additional solutions when csc x = – :  and

. So, the solutions of csc x =  and cot2 x + 1

= 2 are not equivalent.

OPEN ENDED Write a trigonometric equation that has each of the following solutions.

61. 

SOLUTION:  Sample answer: When sin x = 0, x = 0 and x = π.

When sin x = , x =  and x = . So, one

equation that has solutions of 0, π, , and  is 

= 0 or sin2 x –  = 0. This

can be rewritten as 2 sin2 x =  sin x.

62. 

SOLUTION:  

Sample answer: When cos x = , x =  and x =

. When cos x = – , x =  and x = .

So, one equation that has solutions of , , ,

and  is  = 0 or cos2 x

–  = 0. This can be rewritten as 4 cos2 x = 3.

63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying identities.

SOLUTION:  When solving an equation, you use properties of equality to manipulate each side of the equation to isolate a variable. For example:

  When verifying an identity, you transform an expression on one side of the identity into the expression on the other side through a series of algebraic steps.For example:

Verify = .  

Verify each identity.

64.  =

SOLUTION:  

65.  =

SOLUTION:  

66.  +  = 1

SOLUTION:  

+  

= cos2 θ + sin2 θ            Reciprocal Identities= 1                               Pythagorean Identity

Find the value of each expression using the given information.

67. tan ; sin θ = , tan  > 0

SOLUTION:  Use the Pythagorean Identity that involves sin θ.

Since tan  is positive and sin  is positive, cos θ

must be positive. So, .

Solve each equation for all values of x.1. 5 sin x + 2 = sin x

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is  + 2nπ,  + 2nπ, .

2. 5 =  sec 2 x + 3

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

3. 2 = 4 cos2 x + 1

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on

the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

4. 4 tan x – 7 = 3 tan x – 6

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

5. 9 + cot2 x = 12

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

6. 2 – 10 sec x = 4 – 9 sec x

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

7. 3 csc x = 2 csc x +

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

8. 11 =  3 csc2 x + 7

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are , , , and . Solutions

on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

9. 6 tan2 x – 2 = 4

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

10. 9 + sin2 x = 10

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the interval

(– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the solutions is

 + 2nπ,  + 2nπ, .

11. 7 cot x –   = 4 cot x

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

12. 7 cos x = 5 cos x +

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ,   + 2nπ, .

Find all solutions of each equation on [0, 2 ).13. sin4 x + 2 sin2 x − 3 = 0

SOLUTION:  

On the interval [0, 2π),  when x =  and 

 when x = . Since  is not a real 

number, the equation  yields no additional solutions.

14. –2 sin x = –sin x cos x

SOLUTION:  

The equation cos x = 2 has no real solutions since the maximum value the cosine function can obtain is 1. On the interval [0, 2π), the equation  has solutions 0 and π.

15. 4 cot x = cot x sin2 x

SOLUTION:  

The equations sin x = 2 and sin x = –2 have no real solutions. On the interval [0, 2π), the equation cot x =

0 has solutions  and  .

16. csc2 x – csc x + 9 = 11

SOLUTION:  

On the interval [0, 2π), the equation csc x = –1 has a

solution of  and the equation csc x = 2 has

solutions of    and  .

17. cos3 x + cos2 x – cos x = 1

SOLUTION:  

On the interval [0, 2π), the equation cos x = 1 has a solution of 0 and the equation cos x = –1 has a solution of  π.

18. 2 sin2 x = sin x + 1

SOLUTION:  

On the interval [0, 2π), the equation sin x = 1 has a

solution of  and the equation sin x =  has  

solutions of    and  .

19. TENNIS A tennis ball leaves a racquet and heads toward a net 40 feet away. The height of the net is the same height as the initial height of the tennis ball. 

a. If the ball is hit at 50 feet per second, neglecting air resistance, use  to find the interval of possible angles of the ball needed to clear the net. b. Find  if the initial velocity remained the same but the distance to the net was 50 feet.

SOLUTION:  a.

The interval is [15.4°, 74.6°].   b.

If the distance to the net is 50 feet, then the angle would be 19.9° or 70.1°.

20. SKIING In the Olympics aerial skiing competition, skiers speed down a slope that launches them into the air, as shown. The maximum height a skier can

reach is given by hpeak =  , where g is

the acceleration due to gravity or 9.8 meters per second squared.

a. If a skier obtains a height of 5 meters above the end of the ramp, what was the skier’s initial speed? b. Use your answer from part a to determine how long it took the skier to reach the maximum height if tpeak = .

SOLUTION:  a.

The skier’s initial speed was 11.67 meters per second.   b.

  The time it took the skier to reach the maximum height was about 1.0 seconds.

Find all solutions of each equation on the interval [0, 2 ).

21. 1 = cot 2 x + csc x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

22. sec x = tan x + 1

SOLUTION:  

   

  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is 0.

23. tan 2 x = 1 – sec x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are 0,

, and .

24. csc x + cot x = 1

SOLUTION:  

 

x =  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is .

25. 2 – 2 cos2 x = sin x + 1

SOLUTION:  

Check

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

26. cos x – 4 = sin x – 4

SOLUTION:   

 

Therefore, on the interval [0, 2π) the only valid

solutions are and .

27. 3 sin x = 3 – 3 cos x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the only valid

solutions are and 0.

28. cot2 x csc2 x – cot2 x = 9

SOLUTION:  

is undefined. The solutions of  are 

  Check

 

 

 

On the interval , there are no real values of x for which , but  for 

29. sec2 x – 1 + tan x –  tan x =

SOLUTION:  

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

30. sec2 x tan2 x + 3 sec2 x – 2 tan2 x = 3

SOLUTION:  

 

The square of any real number must be greater than or equal to zero so sec2 x = –1 has no solutions. Therefore, on the interval [0, 2π) the only solutions are 0 and π.

31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive power PR of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical components, PH and PV.

Using the equations above, determine for what values of  PV and PH are equivalent.

SOLUTION:  

The sine and cosine have the same values in the

interval [0, 2π) at  and  .

Therefore, the components will be equivalent when

.

Find all solutions of each equation on the interval [0, 2 ).

32.  + cos x = 2

SOLUTION:  

On the interval [0, 2π), cos x =  when x =  and 

when x = .

33.  +  = –4

SOLUTION:  

On the interval [0, 2π),cos x = – when x =

 and when x = .

34.  +  = cos x

SOLUTION:  

On the interval [0, 2π) there are no values of x for

which sin x = –2 and the sin x = 1 only when x = .

However, tan  is undefined, so there are no 

solutions for the original equation.

35. cot x cos x + 1 =  + 

SOLUTION:  

On ,  when x =  and when x =

.

GRAPHING CALCULATOR Solve each equation on the interval [0, 2 ) by graphing. Round to the nearest hundredth.

36. 3 cos 2x = ex + 1

SOLUTION:  

On the interval , the only solution is when x = 0.33.

37. sin x + cos x = 3x

SOLUTION:  

On the interval , the only solution is when x = 0.41.

38. x2 = 2 cos x + x

SOLUTION:  

On the interval , the only solution is when x = 1.34.

39. x log x + 5x cos x = –2

SOLUTION:  

On the interval , the solutions are when x = 1.84 and when x = 4.49.

40. METEOROLOGY The average daily temperature in degrees Fahrenheit for a city can be modeled by t

= 8.05 cos + 66.95, where x is a function of

time, x = 1 represents January 15, x = 2 represents February 15, and so on. a. Use a graphing calculator to estimate the temperature on January 31. b. Approximate the number of months that the average daily temperature is greater than 70 throughout the entire month. c. Estimate the highest temperature of the year and the month in which it occurs.

SOLUTION:  a. Use a graphing calculator to graph t = 8.05

cos + 66.95. Because x = 1 represents

January 15, x = 1.5 represents January 31. Use the TRACE feature to determine the temperature whenx = 1.5.

Therefore, the temperature on January 31 is about 61.3º.   b. Graph y = 70. Then use the intersect feature under the CALC menu to determine the interval on

which 8.05 cos + 66.95 > 70.

So, 8.05 cos + 66.95 > 70 on the interval

3.74 < x < 8.26. Because x = 3.74 corresponds to about the middle of April, the temperature does not exceed 70º for the entire month until May. Therefore, the average daily temperature is greater than 70º for the entire month during the months of May, June, and July. Therefore, the average temperature was higher than 70º for three months.   c. Use the maximum feature under the CALC menu to find the maximum value of the function on [0, 12].

The maximum value is 75 when x ≈ 6. Therefore, the maximum average monthly temperature is 75º during the month of June.

Find the x-intercepts of each graph on the interval [0, 2 ).

41. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) cos x = 0 when x =  and x

= .

42. 

 

SOLUTION:  Let y = 0 and solve for x.

  On the interval [0, 2π) sin x = 0 when x = 0 and x = π.

43. 

SOLUTION:  Let y = 0 and solve for x.

 

On the interval [0, 2π) cot x = 1 when x =  and x

= .

44. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) tan x = 0 when x = 0 and x =

π, tan x = –1 when x =  and x = , and tan x

= 1 when x =  and x = .

Find all solutions of each equation on the interval [0, 4 ).

45. 4 tan x = 2 sec2 x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

46. 2 sin2 x + 1 = –3 sin x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, , , and .

47. csc x cot2 x = csc x

SOLUTION:   

On the interval [0, 4π) the solutions are , ,

, , , , , and .

48. sec x + 5 = 2 sec x + 3

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

49. GEOMETRY Consider the circle below.

a. The length s of  is given by s = r(2 ) where 0≤   ≤  . When s = 18 and AB = 14, the radius is r

= . Use a graphing calculator to find the

measure of 2 in radians. b. The area of the shaded region is given by A =

. Use a graphing calculator to find the

radian measure of θ  if the radius is 5 inches and the area is 36 square inches. Round to the nearest hundredth.

SOLUTION:  a. Rewrite the arclength formula using s = 18 and r

= .

On the graphing calculator, find the intersection of Y1 = 18sinθ and Y2 = 14θ.

The value of 2θ = 2(1.1968) or about 2.39 radians.     b. First, substitute into the given area formula and rearrange it.

  Using a graphing calculator, find the intersection of Y1 = 2.88 and Y2 = θ – sin θ.

When the area is 36 square inches and the radius is 5 inches, then the measure of θ is 3.01 radians.

Solve each inequality on the interval [0, 2 ).50. 1 > 2 sin x

SOLUTION:  Graph y = 1 and y = 2 sin x on [0, 2π). Use the intersect feature under the CALC menu to determine on what interval(s) 1 > 2 sin x.

  The graphs intersect at about 0.524 or  and about 

2.618 or . The graph of y = 1 is above the other

graph when x < and when x > . Therefore, 1

> 2 sin x on 0 ≤ x <    or   < x < 2π.

51. 0 < 2 cos x –

SOLUTION:  Graph y = 2 cos x –  on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0 < 2 cos x – .

  The zeros of y = 2 cos x –  are about 0.785 or  and about 5.498 or  . The graph of y = 0 is

below the other graph when 0 ≤ x <  or   < x <

2π. Therefore, 0 < 2 cos x –  on 0 ≤ x <  or   < 

x < 2π.

52.  ≥ 

SOLUTION:  

Graph y = and y =  on [0, 2π). Use

the intersect feature under the CALC menu to

determine on what interval(s) ≥  .

  The graphs intersect at about 1.047 or  and about 

2.094 or . The graph of y = is below the

other graph when   ≤ x ≤  .

Therefore, ≥   on  ≤ x ≤  .

53.  ≤ tan x cot x

SOLUTION:  

Graph y = and y = tan x cot x on [0, 2π).

Use the intersect feature under the CALC menu to

determine on what interval(s) ≤ tan x cot

x.  

  The graphs intersect at about 3.142 or π. The graph of y = tan x cot x is at or below the other graph when 0 ≤ x < 2π.  

Therefore, ≤ tan x cot x on 0 ≤ x < 2π.

54. cos x ≤

SOLUTION:  

Graph y = cos x and y =  on [0, 2π). Use the

intersect feature under the CALC menu to

determine on what interval(s) cos x ≤ .

  The graphs intersect at about 2.618 or  and about

3.665 or . The graph of y = cos x is below the

other graph when  ≤ x ≤  .

 

Therefore, cos x ≤ on  ≤ x ≤  .

55.  sin x − 1 < 0

SOLUTION:  Graph y =  sin x − 1. Use the zero feature underthe CALC menu to determine on what interval(s)

 sin x − 1 < 0.

  The zeros of  sin x − 1 are about 0.785 or  and

about 2.356 or . The graph is below the x-axis

when 0 ≤ x <  or   < x < 2π.

Therefore,  sin x − 1 < 0 on 0 ≤ x <  or   < 

x < 2π.

56. REFRACTION When light travels from one medium to another it bends or refracts, as shown.  

Refraction is described by n2 sinθ1 = n1 sinθ2, wheren1 is the index of refraction of the medium the light

is entering, n2 is the index of refraction of the medium the light is exiting, θ1 is the angle of

incidence, and θ2 is the angle of refraction. a. Find θ2 for each material shown if the angle of incidence is 40º and the index of refraction for air is 1.00.

b. If the angle of incidence is doubled to 80 , will the resulting angles of refraction be twice as large asthose found in part a?

SOLUTION:  a. glass:

  ice:

  plastic:

  water:

  b. Find the angle of refraction for one of the mediums using an angle of incidence of 80 . glass:

θ2 ≈ 25 when θ1 = 40 andθ2 ≈ 40.4 when θ1 = 80 . Therefore, the angle of refraction is not twice as large.

57. ERROR ANALYSIS Vijay and Alicia are solving tan2 x – tan x +  =   tan x. Vijay thinks that

the solutions are x =  + n , x =  + n , x =

 + n , and x =  + n . Alicia thinks that the

solutions are x =  + n and x =  + n . Is

either of them correct? Explain your reasoning.

SOLUTION:  First, solve tan2 x – tan x +  =   tan x.

  On [0, 2π) tan x = 1 when x =  and x =  and 

tan x = when x =  and x = .

  Sample answer: Therefore, Vijay’s solutions are correct; however, they are not stated in the simplest

form. For example, his solutions of x =  + n and

x =  + n could simply be stated as x =  + 

n because when n = 1,  + nπ  is equivalent to

.

CHALLENGE  Solve each equation on the interval [0, 2π].

58. 16 sin5 x + 2 sin x = 12 sin3 x

SOLUTION:  

On [0, 2π) sin x = 0 when x = 0 and x = π, sin x =

when x =  and x = , sin x = when

x =  and x = , sin x =  – when x =  and 

x = , and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are 0, , , , , , , ,

, and .

59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3

SOLUTION:  

On [0, 2 ) sin x = 1 when x = , sin x  = –1 when

x = , sin x =  – when x =  and x = ,

and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are , , , , , and .

60. REASONING Are the solutions of csc x =  andcot2 x + 1 = 2 equivalent? If so, verify your answer algebraically. If not, explain your reasoning.

SOLUTION:  

 

 

The solutions to csc x =  are   and  . Using

a Pythagorean identity, cot2 x + 1 = 2 simplifies to csc2 x = 2 or csc x = ± . Therefore, there are

two additional solutions when csc x = – :  and

. So, the solutions of csc x =  and cot2 x + 1

= 2 are not equivalent.

OPEN ENDED Write a trigonometric equation that has each of the following solutions.

61. 

SOLUTION:  Sample answer: When sin x = 0, x = 0 and x = π.

When sin x = , x =  and x = . So, one

equation that has solutions of 0, π, , and  is 

= 0 or sin2 x –  = 0. This

can be rewritten as 2 sin2 x =  sin x.

62. 

SOLUTION:  

Sample answer: When cos x = , x =  and x =

. When cos x = – , x =  and x = .

So, one equation that has solutions of , , ,

and  is  = 0 or cos2 x

–  = 0. This can be rewritten as 4 cos2 x = 3.

63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying identities.

SOLUTION:  When solving an equation, you use properties of equality to manipulate each side of the equation to isolate a variable. For example:

  When verifying an identity, you transform an expression on one side of the identity into the expression on the other side through a series of algebraic steps.For example:

Verify = .  

Verify each identity.

64.  =

SOLUTION:  

65.  =

SOLUTION:  

66.  +  = 1

SOLUTION:  

+  

= cos2 θ + sin2 θ            Reciprocal Identities= 1                               Pythagorean Identity

Find the value of each expression using the given information.

67. tan ; sin θ = , tan  > 0

SOLUTION:  Use the Pythagorean Identity that involves sin θ.

Since tan  is positive and sin  is positive, cos θ

must be positive. So, .

eSolutions Manual - Powered by Cognero Page 1

5-3 Solving Trigonometric Equations

Page 2: 5-3 Solving Trigonometric Equations the interval , the only solution is when x = 1.34. 39. x log x + 5x cos x = –2 SOLUTION: On the interval , the solutions are when x = 1.84 and

Solve each equation for all values of x.1. 5 sin x + 2 = sin x

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is  + 2nπ,  + 2nπ, .

2. 5 =  sec 2 x + 3

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

3. 2 = 4 cos2 x + 1

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on

the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

4. 4 tan x – 7 = 3 tan x – 6

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

5. 9 + cot2 x = 12

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

6. 2 – 10 sec x = 4 – 9 sec x

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

7. 3 csc x = 2 csc x +

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

8. 11 =  3 csc2 x + 7

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are , , , and . Solutions

on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

9. 6 tan2 x – 2 = 4

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

10. 9 + sin2 x = 10

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the interval

(– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the solutions is

 + 2nπ,  + 2nπ, .

11. 7 cot x –   = 4 cot x

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

12. 7 cos x = 5 cos x +

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ,   + 2nπ, .

Find all solutions of each equation on [0, 2 ).13. sin4 x + 2 sin2 x − 3 = 0

SOLUTION:  

On the interval [0, 2π),  when x =  and 

 when x = . Since  is not a real 

number, the equation  yields no additional solutions.

14. –2 sin x = –sin x cos x

SOLUTION:  

The equation cos x = 2 has no real solutions since the maximum value the cosine function can obtain is 1. On the interval [0, 2π), the equation  has solutions 0 and π.

15. 4 cot x = cot x sin2 x

SOLUTION:  

The equations sin x = 2 and sin x = –2 have no real solutions. On the interval [0, 2π), the equation cot x =

0 has solutions  and  .

16. csc2 x – csc x + 9 = 11

SOLUTION:  

On the interval [0, 2π), the equation csc x = –1 has a

solution of  and the equation csc x = 2 has

solutions of    and  .

17. cos3 x + cos2 x – cos x = 1

SOLUTION:  

On the interval [0, 2π), the equation cos x = 1 has a solution of 0 and the equation cos x = –1 has a solution of  π.

18. 2 sin2 x = sin x + 1

SOLUTION:  

On the interval [0, 2π), the equation sin x = 1 has a

solution of  and the equation sin x =  has  

solutions of    and  .

19. TENNIS A tennis ball leaves a racquet and heads toward a net 40 feet away. The height of the net is the same height as the initial height of the tennis ball. 

a. If the ball is hit at 50 feet per second, neglecting air resistance, use  to find the interval of possible angles of the ball needed to clear the net. b. Find  if the initial velocity remained the same but the distance to the net was 50 feet.

SOLUTION:  a.

The interval is [15.4°, 74.6°].   b.

If the distance to the net is 50 feet, then the angle would be 19.9° or 70.1°.

20. SKIING In the Olympics aerial skiing competition, skiers speed down a slope that launches them into the air, as shown. The maximum height a skier can

reach is given by hpeak =  , where g is

the acceleration due to gravity or 9.8 meters per second squared.

a. If a skier obtains a height of 5 meters above the end of the ramp, what was the skier’s initial speed? b. Use your answer from part a to determine how long it took the skier to reach the maximum height if tpeak = .

SOLUTION:  a.

The skier’s initial speed was 11.67 meters per second.   b.

  The time it took the skier to reach the maximum height was about 1.0 seconds.

Find all solutions of each equation on the interval [0, 2 ).

21. 1 = cot 2 x + csc x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

22. sec x = tan x + 1

SOLUTION:  

   

  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is 0.

23. tan 2 x = 1 – sec x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are 0,

, and .

24. csc x + cot x = 1

SOLUTION:  

 

x =  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is .

25. 2 – 2 cos2 x = sin x + 1

SOLUTION:  

Check

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

26. cos x – 4 = sin x – 4

SOLUTION:   

 

Therefore, on the interval [0, 2π) the only valid

solutions are and .

27. 3 sin x = 3 – 3 cos x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the only valid

solutions are and 0.

28. cot2 x csc2 x – cot2 x = 9

SOLUTION:  

is undefined. The solutions of  are 

  Check

 

 

 

On the interval , there are no real values of x for which , but  for 

29. sec2 x – 1 + tan x –  tan x =

SOLUTION:  

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

30. sec2 x tan2 x + 3 sec2 x – 2 tan2 x = 3

SOLUTION:  

 

The square of any real number must be greater than or equal to zero so sec2 x = –1 has no solutions. Therefore, on the interval [0, 2π) the only solutions are 0 and π.

31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive power PR of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical components, PH and PV.

Using the equations above, determine for what values of  PV and PH are equivalent.

SOLUTION:  

The sine and cosine have the same values in the

interval [0, 2π) at  and  .

Therefore, the components will be equivalent when

.

Find all solutions of each equation on the interval [0, 2 ).

32.  + cos x = 2

SOLUTION:  

On the interval [0, 2π), cos x =  when x =  and 

when x = .

33.  +  = –4

SOLUTION:  

On the interval [0, 2π),cos x = – when x =

 and when x = .

34.  +  = cos x

SOLUTION:  

On the interval [0, 2π) there are no values of x for

which sin x = –2 and the sin x = 1 only when x = .

However, tan  is undefined, so there are no 

solutions for the original equation.

35. cot x cos x + 1 =  + 

SOLUTION:  

On ,  when x =  and when x =

.

GRAPHING CALCULATOR Solve each equation on the interval [0, 2 ) by graphing. Round to the nearest hundredth.

36. 3 cos 2x = ex + 1

SOLUTION:  

On the interval , the only solution is when x = 0.33.

37. sin x + cos x = 3x

SOLUTION:  

On the interval , the only solution is when x = 0.41.

38. x2 = 2 cos x + x

SOLUTION:  

On the interval , the only solution is when x = 1.34.

39. x log x + 5x cos x = –2

SOLUTION:  

On the interval , the solutions are when x = 1.84 and when x = 4.49.

40. METEOROLOGY The average daily temperature in degrees Fahrenheit for a city can be modeled by t

= 8.05 cos + 66.95, where x is a function of

time, x = 1 represents January 15, x = 2 represents February 15, and so on. a. Use a graphing calculator to estimate the temperature on January 31. b. Approximate the number of months that the average daily temperature is greater than 70 throughout the entire month. c. Estimate the highest temperature of the year and the month in which it occurs.

SOLUTION:  a. Use a graphing calculator to graph t = 8.05

cos + 66.95. Because x = 1 represents

January 15, x = 1.5 represents January 31. Use the TRACE feature to determine the temperature whenx = 1.5.

Therefore, the temperature on January 31 is about 61.3º.   b. Graph y = 70. Then use the intersect feature under the CALC menu to determine the interval on

which 8.05 cos + 66.95 > 70.

So, 8.05 cos + 66.95 > 70 on the interval

3.74 < x < 8.26. Because x = 3.74 corresponds to about the middle of April, the temperature does not exceed 70º for the entire month until May. Therefore, the average daily temperature is greater than 70º for the entire month during the months of May, June, and July. Therefore, the average temperature was higher than 70º for three months.   c. Use the maximum feature under the CALC menu to find the maximum value of the function on [0, 12].

The maximum value is 75 when x ≈ 6. Therefore, the maximum average monthly temperature is 75º during the month of June.

Find the x-intercepts of each graph on the interval [0, 2 ).

41. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) cos x = 0 when x =  and x

= .

42. 

 

SOLUTION:  Let y = 0 and solve for x.

  On the interval [0, 2π) sin x = 0 when x = 0 and x = π.

43. 

SOLUTION:  Let y = 0 and solve for x.

 

On the interval [0, 2π) cot x = 1 when x =  and x

= .

44. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) tan x = 0 when x = 0 and x =

π, tan x = –1 when x =  and x = , and tan x

= 1 when x =  and x = .

Find all solutions of each equation on the interval [0, 4 ).

45. 4 tan x = 2 sec2 x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

46. 2 sin2 x + 1 = –3 sin x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, , , and .

47. csc x cot2 x = csc x

SOLUTION:   

On the interval [0, 4π) the solutions are , ,

, , , , , and .

48. sec x + 5 = 2 sec x + 3

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

49. GEOMETRY Consider the circle below.

a. The length s of  is given by s = r(2 ) where 0≤   ≤  . When s = 18 and AB = 14, the radius is r

= . Use a graphing calculator to find the

measure of 2 in radians. b. The area of the shaded region is given by A =

. Use a graphing calculator to find the

radian measure of θ  if the radius is 5 inches and the area is 36 square inches. Round to the nearest hundredth.

SOLUTION:  a. Rewrite the arclength formula using s = 18 and r

= .

On the graphing calculator, find the intersection of Y1 = 18sinθ and Y2 = 14θ.

The value of 2θ = 2(1.1968) or about 2.39 radians.     b. First, substitute into the given area formula and rearrange it.

  Using a graphing calculator, find the intersection of Y1 = 2.88 and Y2 = θ – sin θ.

When the area is 36 square inches and the radius is 5 inches, then the measure of θ is 3.01 radians.

Solve each inequality on the interval [0, 2 ).50. 1 > 2 sin x

SOLUTION:  Graph y = 1 and y = 2 sin x on [0, 2π). Use the intersect feature under the CALC menu to determine on what interval(s) 1 > 2 sin x.

  The graphs intersect at about 0.524 or  and about 

2.618 or . The graph of y = 1 is above the other

graph when x < and when x > . Therefore, 1

> 2 sin x on 0 ≤ x <    or   < x < 2π.

51. 0 < 2 cos x –

SOLUTION:  Graph y = 2 cos x –  on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0 < 2 cos x – .

  The zeros of y = 2 cos x –  are about 0.785 or  and about 5.498 or  . The graph of y = 0 is

below the other graph when 0 ≤ x <  or   < x <

2π. Therefore, 0 < 2 cos x –  on 0 ≤ x <  or   < 

x < 2π.

52.  ≥ 

SOLUTION:  

Graph y = and y =  on [0, 2π). Use

the intersect feature under the CALC menu to

determine on what interval(s) ≥  .

  The graphs intersect at about 1.047 or  and about 

2.094 or . The graph of y = is below the

other graph when   ≤ x ≤  .

Therefore, ≥   on  ≤ x ≤  .

53.  ≤ tan x cot x

SOLUTION:  

Graph y = and y = tan x cot x on [0, 2π).

Use the intersect feature under the CALC menu to

determine on what interval(s) ≤ tan x cot

x.  

  The graphs intersect at about 3.142 or π. The graph of y = tan x cot x is at or below the other graph when 0 ≤ x < 2π.  

Therefore, ≤ tan x cot x on 0 ≤ x < 2π.

54. cos x ≤

SOLUTION:  

Graph y = cos x and y =  on [0, 2π). Use the

intersect feature under the CALC menu to

determine on what interval(s) cos x ≤ .

  The graphs intersect at about 2.618 or  and about

3.665 or . The graph of y = cos x is below the

other graph when  ≤ x ≤  .

 

Therefore, cos x ≤ on  ≤ x ≤  .

55.  sin x − 1 < 0

SOLUTION:  Graph y =  sin x − 1. Use the zero feature underthe CALC menu to determine on what interval(s)

 sin x − 1 < 0.

  The zeros of  sin x − 1 are about 0.785 or  and

about 2.356 or . The graph is below the x-axis

when 0 ≤ x <  or   < x < 2π.

Therefore,  sin x − 1 < 0 on 0 ≤ x <  or   < 

x < 2π.

56. REFRACTION When light travels from one medium to another it bends or refracts, as shown.  

Refraction is described by n2 sinθ1 = n1 sinθ2, wheren1 is the index of refraction of the medium the light

is entering, n2 is the index of refraction of the medium the light is exiting, θ1 is the angle of

incidence, and θ2 is the angle of refraction. a. Find θ2 for each material shown if the angle of incidence is 40º and the index of refraction for air is 1.00.

b. If the angle of incidence is doubled to 80 , will the resulting angles of refraction be twice as large asthose found in part a?

SOLUTION:  a. glass:

  ice:

  plastic:

  water:

  b. Find the angle of refraction for one of the mediums using an angle of incidence of 80 . glass:

θ2 ≈ 25 when θ1 = 40 andθ2 ≈ 40.4 when θ1 = 80 . Therefore, the angle of refraction is not twice as large.

57. ERROR ANALYSIS Vijay and Alicia are solving tan2 x – tan x +  =   tan x. Vijay thinks that

the solutions are x =  + n , x =  + n , x =

 + n , and x =  + n . Alicia thinks that the

solutions are x =  + n and x =  + n . Is

either of them correct? Explain your reasoning.

SOLUTION:  First, solve tan2 x – tan x +  =   tan x.

  On [0, 2π) tan x = 1 when x =  and x =  and 

tan x = when x =  and x = .

  Sample answer: Therefore, Vijay’s solutions are correct; however, they are not stated in the simplest

form. For example, his solutions of x =  + n and

x =  + n could simply be stated as x =  + 

n because when n = 1,  + nπ  is equivalent to

.

CHALLENGE  Solve each equation on the interval [0, 2π].

58. 16 sin5 x + 2 sin x = 12 sin3 x

SOLUTION:  

On [0, 2π) sin x = 0 when x = 0 and x = π, sin x =

when x =  and x = , sin x = when

x =  and x = , sin x =  – when x =  and 

x = , and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are 0, , , , , , , ,

, and .

59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3

SOLUTION:  

On [0, 2 ) sin x = 1 when x = , sin x  = –1 when

x = , sin x =  – when x =  and x = ,

and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are , , , , , and .

60. REASONING Are the solutions of csc x =  andcot2 x + 1 = 2 equivalent? If so, verify your answer algebraically. If not, explain your reasoning.

SOLUTION:  

 

 

The solutions to csc x =  are   and  . Using

a Pythagorean identity, cot2 x + 1 = 2 simplifies to csc2 x = 2 or csc x = ± . Therefore, there are

two additional solutions when csc x = – :  and

. So, the solutions of csc x =  and cot2 x + 1

= 2 are not equivalent.

OPEN ENDED Write a trigonometric equation that has each of the following solutions.

61. 

SOLUTION:  Sample answer: When sin x = 0, x = 0 and x = π.

When sin x = , x =  and x = . So, one

equation that has solutions of 0, π, , and  is 

= 0 or sin2 x –  = 0. This

can be rewritten as 2 sin2 x =  sin x.

62. 

SOLUTION:  

Sample answer: When cos x = , x =  and x =

. When cos x = – , x =  and x = .

So, one equation that has solutions of , , ,

and  is  = 0 or cos2 x

–  = 0. This can be rewritten as 4 cos2 x = 3.

63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying identities.

SOLUTION:  When solving an equation, you use properties of equality to manipulate each side of the equation to isolate a variable. For example:

  When verifying an identity, you transform an expression on one side of the identity into the expression on the other side through a series of algebraic steps.For example:

Verify = .  

Verify each identity.

64.  =

SOLUTION:  

65.  =

SOLUTION:  

66.  +  = 1

SOLUTION:  

+  

= cos2 θ + sin2 θ            Reciprocal Identities= 1                               Pythagorean Identity

Find the value of each expression using the given information.

67. tan ; sin θ = , tan  > 0

SOLUTION:  Use the Pythagorean Identity that involves sin θ.

Since tan  is positive and sin  is positive, cos θ

must be positive. So, .

Solve each equation for all values of x.1. 5 sin x + 2 = sin x

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is  + 2nπ,  + 2nπ, .

2. 5 =  sec 2 x + 3

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

3. 2 = 4 cos2 x + 1

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on

the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

4. 4 tan x – 7 = 3 tan x – 6

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

5. 9 + cot2 x = 12

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

6. 2 – 10 sec x = 4 – 9 sec x

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

7. 3 csc x = 2 csc x +

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

8. 11 =  3 csc2 x + 7

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are , , , and . Solutions

on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

9. 6 tan2 x – 2 = 4

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

10. 9 + sin2 x = 10

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the interval

(– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the solutions is

 + 2nπ,  + 2nπ, .

11. 7 cot x –   = 4 cot x

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

12. 7 cos x = 5 cos x +

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ,   + 2nπ, .

Find all solutions of each equation on [0, 2 ).13. sin4 x + 2 sin2 x − 3 = 0

SOLUTION:  

On the interval [0, 2π),  when x =  and 

 when x = . Since  is not a real 

number, the equation  yields no additional solutions.

14. –2 sin x = –sin x cos x

SOLUTION:  

The equation cos x = 2 has no real solutions since the maximum value the cosine function can obtain is 1. On the interval [0, 2π), the equation  has solutions 0 and π.

15. 4 cot x = cot x sin2 x

SOLUTION:  

The equations sin x = 2 and sin x = –2 have no real solutions. On the interval [0, 2π), the equation cot x =

0 has solutions  and  .

16. csc2 x – csc x + 9 = 11

SOLUTION:  

On the interval [0, 2π), the equation csc x = –1 has a

solution of  and the equation csc x = 2 has

solutions of    and  .

17. cos3 x + cos2 x – cos x = 1

SOLUTION:  

On the interval [0, 2π), the equation cos x = 1 has a solution of 0 and the equation cos x = –1 has a solution of  π.

18. 2 sin2 x = sin x + 1

SOLUTION:  

On the interval [0, 2π), the equation sin x = 1 has a

solution of  and the equation sin x =  has  

solutions of    and  .

19. TENNIS A tennis ball leaves a racquet and heads toward a net 40 feet away. The height of the net is the same height as the initial height of the tennis ball. 

a. If the ball is hit at 50 feet per second, neglecting air resistance, use  to find the interval of possible angles of the ball needed to clear the net. b. Find  if the initial velocity remained the same but the distance to the net was 50 feet.

SOLUTION:  a.

The interval is [15.4°, 74.6°].   b.

If the distance to the net is 50 feet, then the angle would be 19.9° or 70.1°.

20. SKIING In the Olympics aerial skiing competition, skiers speed down a slope that launches them into the air, as shown. The maximum height a skier can

reach is given by hpeak =  , where g is

the acceleration due to gravity or 9.8 meters per second squared.

a. If a skier obtains a height of 5 meters above the end of the ramp, what was the skier’s initial speed? b. Use your answer from part a to determine how long it took the skier to reach the maximum height if tpeak = .

SOLUTION:  a.

The skier’s initial speed was 11.67 meters per second.   b.

  The time it took the skier to reach the maximum height was about 1.0 seconds.

Find all solutions of each equation on the interval [0, 2 ).

21. 1 = cot 2 x + csc x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

22. sec x = tan x + 1

SOLUTION:  

   

  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is 0.

23. tan 2 x = 1 – sec x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are 0,

, and .

24. csc x + cot x = 1

SOLUTION:  

 

x =  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is .

25. 2 – 2 cos2 x = sin x + 1

SOLUTION:  

Check

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

26. cos x – 4 = sin x – 4

SOLUTION:   

 

Therefore, on the interval [0, 2π) the only valid

solutions are and .

27. 3 sin x = 3 – 3 cos x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the only valid

solutions are and 0.

28. cot2 x csc2 x – cot2 x = 9

SOLUTION:  

is undefined. The solutions of  are 

  Check

 

 

 

On the interval , there are no real values of x for which , but  for 

29. sec2 x – 1 + tan x –  tan x =

SOLUTION:  

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

30. sec2 x tan2 x + 3 sec2 x – 2 tan2 x = 3

SOLUTION:  

 

The square of any real number must be greater than or equal to zero so sec2 x = –1 has no solutions. Therefore, on the interval [0, 2π) the only solutions are 0 and π.

31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive power PR of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical components, PH and PV.

Using the equations above, determine for what values of  PV and PH are equivalent.

SOLUTION:  

The sine and cosine have the same values in the

interval [0, 2π) at  and  .

Therefore, the components will be equivalent when

.

Find all solutions of each equation on the interval [0, 2 ).

32.  + cos x = 2

SOLUTION:  

On the interval [0, 2π), cos x =  when x =  and 

when x = .

33.  +  = –4

SOLUTION:  

On the interval [0, 2π),cos x = – when x =

 and when x = .

34.  +  = cos x

SOLUTION:  

On the interval [0, 2π) there are no values of x for

which sin x = –2 and the sin x = 1 only when x = .

However, tan  is undefined, so there are no 

solutions for the original equation.

35. cot x cos x + 1 =  + 

SOLUTION:  

On ,  when x =  and when x =

.

GRAPHING CALCULATOR Solve each equation on the interval [0, 2 ) by graphing. Round to the nearest hundredth.

36. 3 cos 2x = ex + 1

SOLUTION:  

On the interval , the only solution is when x = 0.33.

37. sin x + cos x = 3x

SOLUTION:  

On the interval , the only solution is when x = 0.41.

38. x2 = 2 cos x + x

SOLUTION:  

On the interval , the only solution is when x = 1.34.

39. x log x + 5x cos x = –2

SOLUTION:  

On the interval , the solutions are when x = 1.84 and when x = 4.49.

40. METEOROLOGY The average daily temperature in degrees Fahrenheit for a city can be modeled by t

= 8.05 cos + 66.95, where x is a function of

time, x = 1 represents January 15, x = 2 represents February 15, and so on. a. Use a graphing calculator to estimate the temperature on January 31. b. Approximate the number of months that the average daily temperature is greater than 70 throughout the entire month. c. Estimate the highest temperature of the year and the month in which it occurs.

SOLUTION:  a. Use a graphing calculator to graph t = 8.05

cos + 66.95. Because x = 1 represents

January 15, x = 1.5 represents January 31. Use the TRACE feature to determine the temperature whenx = 1.5.

Therefore, the temperature on January 31 is about 61.3º.   b. Graph y = 70. Then use the intersect feature under the CALC menu to determine the interval on

which 8.05 cos + 66.95 > 70.

So, 8.05 cos + 66.95 > 70 on the interval

3.74 < x < 8.26. Because x = 3.74 corresponds to about the middle of April, the temperature does not exceed 70º for the entire month until May. Therefore, the average daily temperature is greater than 70º for the entire month during the months of May, June, and July. Therefore, the average temperature was higher than 70º for three months.   c. Use the maximum feature under the CALC menu to find the maximum value of the function on [0, 12].

The maximum value is 75 when x ≈ 6. Therefore, the maximum average monthly temperature is 75º during the month of June.

Find the x-intercepts of each graph on the interval [0, 2 ).

41. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) cos x = 0 when x =  and x

= .

42. 

 

SOLUTION:  Let y = 0 and solve for x.

  On the interval [0, 2π) sin x = 0 when x = 0 and x = π.

43. 

SOLUTION:  Let y = 0 and solve for x.

 

On the interval [0, 2π) cot x = 1 when x =  and x

= .

44. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) tan x = 0 when x = 0 and x =

π, tan x = –1 when x =  and x = , and tan x

= 1 when x =  and x = .

Find all solutions of each equation on the interval [0, 4 ).

45. 4 tan x = 2 sec2 x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

46. 2 sin2 x + 1 = –3 sin x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, , , and .

47. csc x cot2 x = csc x

SOLUTION:   

On the interval [0, 4π) the solutions are , ,

, , , , , and .

48. sec x + 5 = 2 sec x + 3

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

49. GEOMETRY Consider the circle below.

a. The length s of  is given by s = r(2 ) where 0≤   ≤  . When s = 18 and AB = 14, the radius is r

= . Use a graphing calculator to find the

measure of 2 in radians. b. The area of the shaded region is given by A =

. Use a graphing calculator to find the

radian measure of θ  if the radius is 5 inches and the area is 36 square inches. Round to the nearest hundredth.

SOLUTION:  a. Rewrite the arclength formula using s = 18 and r

= .

On the graphing calculator, find the intersection of Y1 = 18sinθ and Y2 = 14θ.

The value of 2θ = 2(1.1968) or about 2.39 radians.     b. First, substitute into the given area formula and rearrange it.

  Using a graphing calculator, find the intersection of Y1 = 2.88 and Y2 = θ – sin θ.

When the area is 36 square inches and the radius is 5 inches, then the measure of θ is 3.01 radians.

Solve each inequality on the interval [0, 2 ).50. 1 > 2 sin x

SOLUTION:  Graph y = 1 and y = 2 sin x on [0, 2π). Use the intersect feature under the CALC menu to determine on what interval(s) 1 > 2 sin x.

  The graphs intersect at about 0.524 or  and about 

2.618 or . The graph of y = 1 is above the other

graph when x < and when x > . Therefore, 1

> 2 sin x on 0 ≤ x <    or   < x < 2π.

51. 0 < 2 cos x –

SOLUTION:  Graph y = 2 cos x –  on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0 < 2 cos x – .

  The zeros of y = 2 cos x –  are about 0.785 or  and about 5.498 or  . The graph of y = 0 is

below the other graph when 0 ≤ x <  or   < x <

2π. Therefore, 0 < 2 cos x –  on 0 ≤ x <  or   < 

x < 2π.

52.  ≥ 

SOLUTION:  

Graph y = and y =  on [0, 2π). Use

the intersect feature under the CALC menu to

determine on what interval(s) ≥  .

  The graphs intersect at about 1.047 or  and about 

2.094 or . The graph of y = is below the

other graph when   ≤ x ≤  .

Therefore, ≥   on  ≤ x ≤  .

53.  ≤ tan x cot x

SOLUTION:  

Graph y = and y = tan x cot x on [0, 2π).

Use the intersect feature under the CALC menu to

determine on what interval(s) ≤ tan x cot

x.  

  The graphs intersect at about 3.142 or π. The graph of y = tan x cot x is at or below the other graph when 0 ≤ x < 2π.  

Therefore, ≤ tan x cot x on 0 ≤ x < 2π.

54. cos x ≤

SOLUTION:  

Graph y = cos x and y =  on [0, 2π). Use the

intersect feature under the CALC menu to

determine on what interval(s) cos x ≤ .

  The graphs intersect at about 2.618 or  and about

3.665 or . The graph of y = cos x is below the

other graph when  ≤ x ≤  .

 

Therefore, cos x ≤ on  ≤ x ≤  .

55.  sin x − 1 < 0

SOLUTION:  Graph y =  sin x − 1. Use the zero feature underthe CALC menu to determine on what interval(s)

 sin x − 1 < 0.

  The zeros of  sin x − 1 are about 0.785 or  and

about 2.356 or . The graph is below the x-axis

when 0 ≤ x <  or   < x < 2π.

Therefore,  sin x − 1 < 0 on 0 ≤ x <  or   < 

x < 2π.

56. REFRACTION When light travels from one medium to another it bends or refracts, as shown.  

Refraction is described by n2 sinθ1 = n1 sinθ2, wheren1 is the index of refraction of the medium the light

is entering, n2 is the index of refraction of the medium the light is exiting, θ1 is the angle of

incidence, and θ2 is the angle of refraction. a. Find θ2 for each material shown if the angle of incidence is 40º and the index of refraction for air is 1.00.

b. If the angle of incidence is doubled to 80 , will the resulting angles of refraction be twice as large asthose found in part a?

SOLUTION:  a. glass:

  ice:

  plastic:

  water:

  b. Find the angle of refraction for one of the mediums using an angle of incidence of 80 . glass:

θ2 ≈ 25 when θ1 = 40 andθ2 ≈ 40.4 when θ1 = 80 . Therefore, the angle of refraction is not twice as large.

57. ERROR ANALYSIS Vijay and Alicia are solving tan2 x – tan x +  =   tan x. Vijay thinks that

the solutions are x =  + n , x =  + n , x =

 + n , and x =  + n . Alicia thinks that the

solutions are x =  + n and x =  + n . Is

either of them correct? Explain your reasoning.

SOLUTION:  First, solve tan2 x – tan x +  =   tan x.

  On [0, 2π) tan x = 1 when x =  and x =  and 

tan x = when x =  and x = .

  Sample answer: Therefore, Vijay’s solutions are correct; however, they are not stated in the simplest

form. For example, his solutions of x =  + n and

x =  + n could simply be stated as x =  + 

n because when n = 1,  + nπ  is equivalent to

.

CHALLENGE  Solve each equation on the interval [0, 2π].

58. 16 sin5 x + 2 sin x = 12 sin3 x

SOLUTION:  

On [0, 2π) sin x = 0 when x = 0 and x = π, sin x =

when x =  and x = , sin x = when

x =  and x = , sin x =  – when x =  and 

x = , and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are 0, , , , , , , ,

, and .

59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3

SOLUTION:  

On [0, 2 ) sin x = 1 when x = , sin x  = –1 when

x = , sin x =  – when x =  and x = ,

and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are , , , , , and .

60. REASONING Are the solutions of csc x =  andcot2 x + 1 = 2 equivalent? If so, verify your answer algebraically. If not, explain your reasoning.

SOLUTION:  

 

 

The solutions to csc x =  are   and  . Using

a Pythagorean identity, cot2 x + 1 = 2 simplifies to csc2 x = 2 or csc x = ± . Therefore, there are

two additional solutions when csc x = – :  and

. So, the solutions of csc x =  and cot2 x + 1

= 2 are not equivalent.

OPEN ENDED Write a trigonometric equation that has each of the following solutions.

61. 

SOLUTION:  Sample answer: When sin x = 0, x = 0 and x = π.

When sin x = , x =  and x = . So, one

equation that has solutions of 0, π, , and  is 

= 0 or sin2 x –  = 0. This

can be rewritten as 2 sin2 x =  sin x.

62. 

SOLUTION:  

Sample answer: When cos x = , x =  and x =

. When cos x = – , x =  and x = .

So, one equation that has solutions of , , ,

and  is  = 0 or cos2 x

–  = 0. This can be rewritten as 4 cos2 x = 3.

63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying identities.

SOLUTION:  When solving an equation, you use properties of equality to manipulate each side of the equation to isolate a variable. For example:

  When verifying an identity, you transform an expression on one side of the identity into the expression on the other side through a series of algebraic steps.For example:

Verify = .  

Verify each identity.

64.  =

SOLUTION:  

65.  =

SOLUTION:  

66.  +  = 1

SOLUTION:  

+  

= cos2 θ + sin2 θ            Reciprocal Identities= 1                               Pythagorean Identity

Find the value of each expression using the given information.

67. tan ; sin θ = , tan  > 0

SOLUTION:  Use the Pythagorean Identity that involves sin θ.

Since tan  is positive and sin  is positive, cos θ

must be positive. So, .

eSolutions Manual - Powered by Cognero Page 2

5-3 Solving Trigonometric Equations

Page 3: 5-3 Solving Trigonometric Equations the interval , the only solution is when x = 1.34. 39. x log x + 5x cos x = –2 SOLUTION: On the interval , the solutions are when x = 1.84 and

Solve each equation for all values of x.1. 5 sin x + 2 = sin x

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is  + 2nπ,  + 2nπ, .

2. 5 =  sec 2 x + 3

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

3. 2 = 4 cos2 x + 1

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on

the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

4. 4 tan x – 7 = 3 tan x – 6

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

5. 9 + cot2 x = 12

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

6. 2 – 10 sec x = 4 – 9 sec x

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

7. 3 csc x = 2 csc x +

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

8. 11 =  3 csc2 x + 7

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are , , , and . Solutions

on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

9. 6 tan2 x – 2 = 4

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

10. 9 + sin2 x = 10

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the interval

(– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the solutions is

 + 2nπ,  + 2nπ, .

11. 7 cot x –   = 4 cot x

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

12. 7 cos x = 5 cos x +

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ,   + 2nπ, .

Find all solutions of each equation on [0, 2 ).13. sin4 x + 2 sin2 x − 3 = 0

SOLUTION:  

On the interval [0, 2π),  when x =  and 

 when x = . Since  is not a real 

number, the equation  yields no additional solutions.

14. –2 sin x = –sin x cos x

SOLUTION:  

The equation cos x = 2 has no real solutions since the maximum value the cosine function can obtain is 1. On the interval [0, 2π), the equation  has solutions 0 and π.

15. 4 cot x = cot x sin2 x

SOLUTION:  

The equations sin x = 2 and sin x = –2 have no real solutions. On the interval [0, 2π), the equation cot x =

0 has solutions  and  .

16. csc2 x – csc x + 9 = 11

SOLUTION:  

On the interval [0, 2π), the equation csc x = –1 has a

solution of  and the equation csc x = 2 has

solutions of    and  .

17. cos3 x + cos2 x – cos x = 1

SOLUTION:  

On the interval [0, 2π), the equation cos x = 1 has a solution of 0 and the equation cos x = –1 has a solution of  π.

18. 2 sin2 x = sin x + 1

SOLUTION:  

On the interval [0, 2π), the equation sin x = 1 has a

solution of  and the equation sin x =  has  

solutions of    and  .

19. TENNIS A tennis ball leaves a racquet and heads toward a net 40 feet away. The height of the net is the same height as the initial height of the tennis ball. 

a. If the ball is hit at 50 feet per second, neglecting air resistance, use  to find the interval of possible angles of the ball needed to clear the net. b. Find  if the initial velocity remained the same but the distance to the net was 50 feet.

SOLUTION:  a.

The interval is [15.4°, 74.6°].   b.

If the distance to the net is 50 feet, then the angle would be 19.9° or 70.1°.

20. SKIING In the Olympics aerial skiing competition, skiers speed down a slope that launches them into the air, as shown. The maximum height a skier can

reach is given by hpeak =  , where g is

the acceleration due to gravity or 9.8 meters per second squared.

a. If a skier obtains a height of 5 meters above the end of the ramp, what was the skier’s initial speed? b. Use your answer from part a to determine how long it took the skier to reach the maximum height if tpeak = .

SOLUTION:  a.

The skier’s initial speed was 11.67 meters per second.   b.

  The time it took the skier to reach the maximum height was about 1.0 seconds.

Find all solutions of each equation on the interval [0, 2 ).

21. 1 = cot 2 x + csc x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

22. sec x = tan x + 1

SOLUTION:  

   

  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is 0.

23. tan 2 x = 1 – sec x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are 0,

, and .

24. csc x + cot x = 1

SOLUTION:  

 

x =  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is .

25. 2 – 2 cos2 x = sin x + 1

SOLUTION:  

Check

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

26. cos x – 4 = sin x – 4

SOLUTION:   

 

Therefore, on the interval [0, 2π) the only valid

solutions are and .

27. 3 sin x = 3 – 3 cos x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the only valid

solutions are and 0.

28. cot2 x csc2 x – cot2 x = 9

SOLUTION:  

is undefined. The solutions of  are 

  Check

 

 

 

On the interval , there are no real values of x for which , but  for 

29. sec2 x – 1 + tan x –  tan x =

SOLUTION:  

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

30. sec2 x tan2 x + 3 sec2 x – 2 tan2 x = 3

SOLUTION:  

 

The square of any real number must be greater than or equal to zero so sec2 x = –1 has no solutions. Therefore, on the interval [0, 2π) the only solutions are 0 and π.

31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive power PR of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical components, PH and PV.

Using the equations above, determine for what values of  PV and PH are equivalent.

SOLUTION:  

The sine and cosine have the same values in the

interval [0, 2π) at  and  .

Therefore, the components will be equivalent when

.

Find all solutions of each equation on the interval [0, 2 ).

32.  + cos x = 2

SOLUTION:  

On the interval [0, 2π), cos x =  when x =  and 

when x = .

33.  +  = –4

SOLUTION:  

On the interval [0, 2π),cos x = – when x =

 and when x = .

34.  +  = cos x

SOLUTION:  

On the interval [0, 2π) there are no values of x for

which sin x = –2 and the sin x = 1 only when x = .

However, tan  is undefined, so there are no 

solutions for the original equation.

35. cot x cos x + 1 =  + 

SOLUTION:  

On ,  when x =  and when x =

.

GRAPHING CALCULATOR Solve each equation on the interval [0, 2 ) by graphing. Round to the nearest hundredth.

36. 3 cos 2x = ex + 1

SOLUTION:  

On the interval , the only solution is when x = 0.33.

37. sin x + cos x = 3x

SOLUTION:  

On the interval , the only solution is when x = 0.41.

38. x2 = 2 cos x + x

SOLUTION:  

On the interval , the only solution is when x = 1.34.

39. x log x + 5x cos x = –2

SOLUTION:  

On the interval , the solutions are when x = 1.84 and when x = 4.49.

40. METEOROLOGY The average daily temperature in degrees Fahrenheit for a city can be modeled by t

= 8.05 cos + 66.95, where x is a function of

time, x = 1 represents January 15, x = 2 represents February 15, and so on. a. Use a graphing calculator to estimate the temperature on January 31. b. Approximate the number of months that the average daily temperature is greater than 70 throughout the entire month. c. Estimate the highest temperature of the year and the month in which it occurs.

SOLUTION:  a. Use a graphing calculator to graph t = 8.05

cos + 66.95. Because x = 1 represents

January 15, x = 1.5 represents January 31. Use the TRACE feature to determine the temperature whenx = 1.5.

Therefore, the temperature on January 31 is about 61.3º.   b. Graph y = 70. Then use the intersect feature under the CALC menu to determine the interval on

which 8.05 cos + 66.95 > 70.

So, 8.05 cos + 66.95 > 70 on the interval

3.74 < x < 8.26. Because x = 3.74 corresponds to about the middle of April, the temperature does not exceed 70º for the entire month until May. Therefore, the average daily temperature is greater than 70º for the entire month during the months of May, June, and July. Therefore, the average temperature was higher than 70º for three months.   c. Use the maximum feature under the CALC menu to find the maximum value of the function on [0, 12].

The maximum value is 75 when x ≈ 6. Therefore, the maximum average monthly temperature is 75º during the month of June.

Find the x-intercepts of each graph on the interval [0, 2 ).

41. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) cos x = 0 when x =  and x

= .

42. 

 

SOLUTION:  Let y = 0 and solve for x.

  On the interval [0, 2π) sin x = 0 when x = 0 and x = π.

43. 

SOLUTION:  Let y = 0 and solve for x.

 

On the interval [0, 2π) cot x = 1 when x =  and x

= .

44. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) tan x = 0 when x = 0 and x =

π, tan x = –1 when x =  and x = , and tan x

= 1 when x =  and x = .

Find all solutions of each equation on the interval [0, 4 ).

45. 4 tan x = 2 sec2 x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

46. 2 sin2 x + 1 = –3 sin x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, , , and .

47. csc x cot2 x = csc x

SOLUTION:   

On the interval [0, 4π) the solutions are , ,

, , , , , and .

48. sec x + 5 = 2 sec x + 3

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

49. GEOMETRY Consider the circle below.

a. The length s of  is given by s = r(2 ) where 0≤   ≤  . When s = 18 and AB = 14, the radius is r

= . Use a graphing calculator to find the

measure of 2 in radians. b. The area of the shaded region is given by A =

. Use a graphing calculator to find the

radian measure of θ  if the radius is 5 inches and the area is 36 square inches. Round to the nearest hundredth.

SOLUTION:  a. Rewrite the arclength formula using s = 18 and r

= .

On the graphing calculator, find the intersection of Y1 = 18sinθ and Y2 = 14θ.

The value of 2θ = 2(1.1968) or about 2.39 radians.     b. First, substitute into the given area formula and rearrange it.

  Using a graphing calculator, find the intersection of Y1 = 2.88 and Y2 = θ – sin θ.

When the area is 36 square inches and the radius is 5 inches, then the measure of θ is 3.01 radians.

Solve each inequality on the interval [0, 2 ).50. 1 > 2 sin x

SOLUTION:  Graph y = 1 and y = 2 sin x on [0, 2π). Use the intersect feature under the CALC menu to determine on what interval(s) 1 > 2 sin x.

  The graphs intersect at about 0.524 or  and about 

2.618 or . The graph of y = 1 is above the other

graph when x < and when x > . Therefore, 1

> 2 sin x on 0 ≤ x <    or   < x < 2π.

51. 0 < 2 cos x –

SOLUTION:  Graph y = 2 cos x –  on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0 < 2 cos x – .

  The zeros of y = 2 cos x –  are about 0.785 or  and about 5.498 or  . The graph of y = 0 is

below the other graph when 0 ≤ x <  or   < x <

2π. Therefore, 0 < 2 cos x –  on 0 ≤ x <  or   < 

x < 2π.

52.  ≥ 

SOLUTION:  

Graph y = and y =  on [0, 2π). Use

the intersect feature under the CALC menu to

determine on what interval(s) ≥  .

  The graphs intersect at about 1.047 or  and about 

2.094 or . The graph of y = is below the

other graph when   ≤ x ≤  .

Therefore, ≥   on  ≤ x ≤  .

53.  ≤ tan x cot x

SOLUTION:  

Graph y = and y = tan x cot x on [0, 2π).

Use the intersect feature under the CALC menu to

determine on what interval(s) ≤ tan x cot

x.  

  The graphs intersect at about 3.142 or π. The graph of y = tan x cot x is at or below the other graph when 0 ≤ x < 2π.  

Therefore, ≤ tan x cot x on 0 ≤ x < 2π.

54. cos x ≤

SOLUTION:  

Graph y = cos x and y =  on [0, 2π). Use the

intersect feature under the CALC menu to

determine on what interval(s) cos x ≤ .

  The graphs intersect at about 2.618 or  and about

3.665 or . The graph of y = cos x is below the

other graph when  ≤ x ≤  .

 

Therefore, cos x ≤ on  ≤ x ≤  .

55.  sin x − 1 < 0

SOLUTION:  Graph y =  sin x − 1. Use the zero feature underthe CALC menu to determine on what interval(s)

 sin x − 1 < 0.

  The zeros of  sin x − 1 are about 0.785 or  and

about 2.356 or . The graph is below the x-axis

when 0 ≤ x <  or   < x < 2π.

Therefore,  sin x − 1 < 0 on 0 ≤ x <  or   < 

x < 2π.

56. REFRACTION When light travels from one medium to another it bends or refracts, as shown.  

Refraction is described by n2 sinθ1 = n1 sinθ2, wheren1 is the index of refraction of the medium the light

is entering, n2 is the index of refraction of the medium the light is exiting, θ1 is the angle of

incidence, and θ2 is the angle of refraction. a. Find θ2 for each material shown if the angle of incidence is 40º and the index of refraction for air is 1.00.

b. If the angle of incidence is doubled to 80 , will the resulting angles of refraction be twice as large asthose found in part a?

SOLUTION:  a. glass:

  ice:

  plastic:

  water:

  b. Find the angle of refraction for one of the mediums using an angle of incidence of 80 . glass:

θ2 ≈ 25 when θ1 = 40 andθ2 ≈ 40.4 when θ1 = 80 . Therefore, the angle of refraction is not twice as large.

57. ERROR ANALYSIS Vijay and Alicia are solving tan2 x – tan x +  =   tan x. Vijay thinks that

the solutions are x =  + n , x =  + n , x =

 + n , and x =  + n . Alicia thinks that the

solutions are x =  + n and x =  + n . Is

either of them correct? Explain your reasoning.

SOLUTION:  First, solve tan2 x – tan x +  =   tan x.

  On [0, 2π) tan x = 1 when x =  and x =  and 

tan x = when x =  and x = .

  Sample answer: Therefore, Vijay’s solutions are correct; however, they are not stated in the simplest

form. For example, his solutions of x =  + n and

x =  + n could simply be stated as x =  + 

n because when n = 1,  + nπ  is equivalent to

.

CHALLENGE  Solve each equation on the interval [0, 2π].

58. 16 sin5 x + 2 sin x = 12 sin3 x

SOLUTION:  

On [0, 2π) sin x = 0 when x = 0 and x = π, sin x =

when x =  and x = , sin x = when

x =  and x = , sin x =  – when x =  and 

x = , and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are 0, , , , , , , ,

, and .

59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3

SOLUTION:  

On [0, 2 ) sin x = 1 when x = , sin x  = –1 when

x = , sin x =  – when x =  and x = ,

and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are , , , , , and .

60. REASONING Are the solutions of csc x =  andcot2 x + 1 = 2 equivalent? If so, verify your answer algebraically. If not, explain your reasoning.

SOLUTION:  

 

 

The solutions to csc x =  are   and  . Using

a Pythagorean identity, cot2 x + 1 = 2 simplifies to csc2 x = 2 or csc x = ± . Therefore, there are

two additional solutions when csc x = – :  and

. So, the solutions of csc x =  and cot2 x + 1

= 2 are not equivalent.

OPEN ENDED Write a trigonometric equation that has each of the following solutions.

61. 

SOLUTION:  Sample answer: When sin x = 0, x = 0 and x = π.

When sin x = , x =  and x = . So, one

equation that has solutions of 0, π, , and  is 

= 0 or sin2 x –  = 0. This

can be rewritten as 2 sin2 x =  sin x.

62. 

SOLUTION:  

Sample answer: When cos x = , x =  and x =

. When cos x = – , x =  and x = .

So, one equation that has solutions of , , ,

and  is  = 0 or cos2 x

–  = 0. This can be rewritten as 4 cos2 x = 3.

63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying identities.

SOLUTION:  When solving an equation, you use properties of equality to manipulate each side of the equation to isolate a variable. For example:

  When verifying an identity, you transform an expression on one side of the identity into the expression on the other side through a series of algebraic steps.For example:

Verify = .  

Verify each identity.

64.  =

SOLUTION:  

65.  =

SOLUTION:  

66.  +  = 1

SOLUTION:  

+  

= cos2 θ + sin2 θ            Reciprocal Identities= 1                               Pythagorean Identity

Find the value of each expression using the given information.

67. tan ; sin θ = , tan  > 0

SOLUTION:  Use the Pythagorean Identity that involves sin θ.

Since tan  is positive and sin  is positive, cos θ

must be positive. So, .

Solve each equation for all values of x.1. 5 sin x + 2 = sin x

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is  + 2nπ,  + 2nπ, .

2. 5 =  sec 2 x + 3

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

3. 2 = 4 cos2 x + 1

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on

the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

4. 4 tan x – 7 = 3 tan x – 6

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

5. 9 + cot2 x = 12

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

6. 2 – 10 sec x = 4 – 9 sec x

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

7. 3 csc x = 2 csc x +

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

8. 11 =  3 csc2 x + 7

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are , , , and . Solutions

on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

9. 6 tan2 x – 2 = 4

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

10. 9 + sin2 x = 10

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the interval

(– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the solutions is

 + 2nπ,  + 2nπ, .

11. 7 cot x –   = 4 cot x

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

12. 7 cos x = 5 cos x +

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ,   + 2nπ, .

Find all solutions of each equation on [0, 2 ).13. sin4 x + 2 sin2 x − 3 = 0

SOLUTION:  

On the interval [0, 2π),  when x =  and 

 when x = . Since  is not a real 

number, the equation  yields no additional solutions.

14. –2 sin x = –sin x cos x

SOLUTION:  

The equation cos x = 2 has no real solutions since the maximum value the cosine function can obtain is 1. On the interval [0, 2π), the equation  has solutions 0 and π.

15. 4 cot x = cot x sin2 x

SOLUTION:  

The equations sin x = 2 and sin x = –2 have no real solutions. On the interval [0, 2π), the equation cot x =

0 has solutions  and  .

16. csc2 x – csc x + 9 = 11

SOLUTION:  

On the interval [0, 2π), the equation csc x = –1 has a

solution of  and the equation csc x = 2 has

solutions of    and  .

17. cos3 x + cos2 x – cos x = 1

SOLUTION:  

On the interval [0, 2π), the equation cos x = 1 has a solution of 0 and the equation cos x = –1 has a solution of  π.

18. 2 sin2 x = sin x + 1

SOLUTION:  

On the interval [0, 2π), the equation sin x = 1 has a

solution of  and the equation sin x =  has  

solutions of    and  .

19. TENNIS A tennis ball leaves a racquet and heads toward a net 40 feet away. The height of the net is the same height as the initial height of the tennis ball. 

a. If the ball is hit at 50 feet per second, neglecting air resistance, use  to find the interval of possible angles of the ball needed to clear the net. b. Find  if the initial velocity remained the same but the distance to the net was 50 feet.

SOLUTION:  a.

The interval is [15.4°, 74.6°].   b.

If the distance to the net is 50 feet, then the angle would be 19.9° or 70.1°.

20. SKIING In the Olympics aerial skiing competition, skiers speed down a slope that launches them into the air, as shown. The maximum height a skier can

reach is given by hpeak =  , where g is

the acceleration due to gravity or 9.8 meters per second squared.

a. If a skier obtains a height of 5 meters above the end of the ramp, what was the skier’s initial speed? b. Use your answer from part a to determine how long it took the skier to reach the maximum height if tpeak = .

SOLUTION:  a.

The skier’s initial speed was 11.67 meters per second.   b.

  The time it took the skier to reach the maximum height was about 1.0 seconds.

Find all solutions of each equation on the interval [0, 2 ).

21. 1 = cot 2 x + csc x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

22. sec x = tan x + 1

SOLUTION:  

   

  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is 0.

23. tan 2 x = 1 – sec x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are 0,

, and .

24. csc x + cot x = 1

SOLUTION:  

 

x =  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is .

25. 2 – 2 cos2 x = sin x + 1

SOLUTION:  

Check

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

26. cos x – 4 = sin x – 4

SOLUTION:   

 

Therefore, on the interval [0, 2π) the only valid

solutions are and .

27. 3 sin x = 3 – 3 cos x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the only valid

solutions are and 0.

28. cot2 x csc2 x – cot2 x = 9

SOLUTION:  

is undefined. The solutions of  are 

  Check

 

 

 

On the interval , there are no real values of x for which , but  for 

29. sec2 x – 1 + tan x –  tan x =

SOLUTION:  

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

30. sec2 x tan2 x + 3 sec2 x – 2 tan2 x = 3

SOLUTION:  

 

The square of any real number must be greater than or equal to zero so sec2 x = –1 has no solutions. Therefore, on the interval [0, 2π) the only solutions are 0 and π.

31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive power PR of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical components, PH and PV.

Using the equations above, determine for what values of  PV and PH are equivalent.

SOLUTION:  

The sine and cosine have the same values in the

interval [0, 2π) at  and  .

Therefore, the components will be equivalent when

.

Find all solutions of each equation on the interval [0, 2 ).

32.  + cos x = 2

SOLUTION:  

On the interval [0, 2π), cos x =  when x =  and 

when x = .

33.  +  = –4

SOLUTION:  

On the interval [0, 2π),cos x = – when x =

 and when x = .

34.  +  = cos x

SOLUTION:  

On the interval [0, 2π) there are no values of x for

which sin x = –2 and the sin x = 1 only when x = .

However, tan  is undefined, so there are no 

solutions for the original equation.

35. cot x cos x + 1 =  + 

SOLUTION:  

On ,  when x =  and when x =

.

GRAPHING CALCULATOR Solve each equation on the interval [0, 2 ) by graphing. Round to the nearest hundredth.

36. 3 cos 2x = ex + 1

SOLUTION:  

On the interval , the only solution is when x = 0.33.

37. sin x + cos x = 3x

SOLUTION:  

On the interval , the only solution is when x = 0.41.

38. x2 = 2 cos x + x

SOLUTION:  

On the interval , the only solution is when x = 1.34.

39. x log x + 5x cos x = –2

SOLUTION:  

On the interval , the solutions are when x = 1.84 and when x = 4.49.

40. METEOROLOGY The average daily temperature in degrees Fahrenheit for a city can be modeled by t

= 8.05 cos + 66.95, where x is a function of

time, x = 1 represents January 15, x = 2 represents February 15, and so on. a. Use a graphing calculator to estimate the temperature on January 31. b. Approximate the number of months that the average daily temperature is greater than 70 throughout the entire month. c. Estimate the highest temperature of the year and the month in which it occurs.

SOLUTION:  a. Use a graphing calculator to graph t = 8.05

cos + 66.95. Because x = 1 represents

January 15, x = 1.5 represents January 31. Use the TRACE feature to determine the temperature whenx = 1.5.

Therefore, the temperature on January 31 is about 61.3º.   b. Graph y = 70. Then use the intersect feature under the CALC menu to determine the interval on

which 8.05 cos + 66.95 > 70.

So, 8.05 cos + 66.95 > 70 on the interval

3.74 < x < 8.26. Because x = 3.74 corresponds to about the middle of April, the temperature does not exceed 70º for the entire month until May. Therefore, the average daily temperature is greater than 70º for the entire month during the months of May, June, and July. Therefore, the average temperature was higher than 70º for three months.   c. Use the maximum feature under the CALC menu to find the maximum value of the function on [0, 12].

The maximum value is 75 when x ≈ 6. Therefore, the maximum average monthly temperature is 75º during the month of June.

Find the x-intercepts of each graph on the interval [0, 2 ).

41. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) cos x = 0 when x =  and x

= .

42. 

 

SOLUTION:  Let y = 0 and solve for x.

  On the interval [0, 2π) sin x = 0 when x = 0 and x = π.

43. 

SOLUTION:  Let y = 0 and solve for x.

 

On the interval [0, 2π) cot x = 1 when x =  and x

= .

44. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) tan x = 0 when x = 0 and x =

π, tan x = –1 when x =  and x = , and tan x

= 1 when x =  and x = .

Find all solutions of each equation on the interval [0, 4 ).

45. 4 tan x = 2 sec2 x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

46. 2 sin2 x + 1 = –3 sin x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, , , and .

47. csc x cot2 x = csc x

SOLUTION:   

On the interval [0, 4π) the solutions are , ,

, , , , , and .

48. sec x + 5 = 2 sec x + 3

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

49. GEOMETRY Consider the circle below.

a. The length s of  is given by s = r(2 ) where 0≤   ≤  . When s = 18 and AB = 14, the radius is r

= . Use a graphing calculator to find the

measure of 2 in radians. b. The area of the shaded region is given by A =

. Use a graphing calculator to find the

radian measure of θ  if the radius is 5 inches and the area is 36 square inches. Round to the nearest hundredth.

SOLUTION:  a. Rewrite the arclength formula using s = 18 and r

= .

On the graphing calculator, find the intersection of Y1 = 18sinθ and Y2 = 14θ.

The value of 2θ = 2(1.1968) or about 2.39 radians.     b. First, substitute into the given area formula and rearrange it.

  Using a graphing calculator, find the intersection of Y1 = 2.88 and Y2 = θ – sin θ.

When the area is 36 square inches and the radius is 5 inches, then the measure of θ is 3.01 radians.

Solve each inequality on the interval [0, 2 ).50. 1 > 2 sin x

SOLUTION:  Graph y = 1 and y = 2 sin x on [0, 2π). Use the intersect feature under the CALC menu to determine on what interval(s) 1 > 2 sin x.

  The graphs intersect at about 0.524 or  and about 

2.618 or . The graph of y = 1 is above the other

graph when x < and when x > . Therefore, 1

> 2 sin x on 0 ≤ x <    or   < x < 2π.

51. 0 < 2 cos x –

SOLUTION:  Graph y = 2 cos x –  on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0 < 2 cos x – .

  The zeros of y = 2 cos x –  are about 0.785 or  and about 5.498 or  . The graph of y = 0 is

below the other graph when 0 ≤ x <  or   < x <

2π. Therefore, 0 < 2 cos x –  on 0 ≤ x <  or   < 

x < 2π.

52.  ≥ 

SOLUTION:  

Graph y = and y =  on [0, 2π). Use

the intersect feature under the CALC menu to

determine on what interval(s) ≥  .

  The graphs intersect at about 1.047 or  and about 

2.094 or . The graph of y = is below the

other graph when   ≤ x ≤  .

Therefore, ≥   on  ≤ x ≤  .

53.  ≤ tan x cot x

SOLUTION:  

Graph y = and y = tan x cot x on [0, 2π).

Use the intersect feature under the CALC menu to

determine on what interval(s) ≤ tan x cot

x.  

  The graphs intersect at about 3.142 or π. The graph of y = tan x cot x is at or below the other graph when 0 ≤ x < 2π.  

Therefore, ≤ tan x cot x on 0 ≤ x < 2π.

54. cos x ≤

SOLUTION:  

Graph y = cos x and y =  on [0, 2π). Use the

intersect feature under the CALC menu to

determine on what interval(s) cos x ≤ .

  The graphs intersect at about 2.618 or  and about

3.665 or . The graph of y = cos x is below the

other graph when  ≤ x ≤  .

 

Therefore, cos x ≤ on  ≤ x ≤  .

55.  sin x − 1 < 0

SOLUTION:  Graph y =  sin x − 1. Use the zero feature underthe CALC menu to determine on what interval(s)

 sin x − 1 < 0.

  The zeros of  sin x − 1 are about 0.785 or  and

about 2.356 or . The graph is below the x-axis

when 0 ≤ x <  or   < x < 2π.

Therefore,  sin x − 1 < 0 on 0 ≤ x <  or   < 

x < 2π.

56. REFRACTION When light travels from one medium to another it bends or refracts, as shown.  

Refraction is described by n2 sinθ1 = n1 sinθ2, wheren1 is the index of refraction of the medium the light

is entering, n2 is the index of refraction of the medium the light is exiting, θ1 is the angle of

incidence, and θ2 is the angle of refraction. a. Find θ2 for each material shown if the angle of incidence is 40º and the index of refraction for air is 1.00.

b. If the angle of incidence is doubled to 80 , will the resulting angles of refraction be twice as large asthose found in part a?

SOLUTION:  a. glass:

  ice:

  plastic:

  water:

  b. Find the angle of refraction for one of the mediums using an angle of incidence of 80 . glass:

θ2 ≈ 25 when θ1 = 40 andθ2 ≈ 40.4 when θ1 = 80 . Therefore, the angle of refraction is not twice as large.

57. ERROR ANALYSIS Vijay and Alicia are solving tan2 x – tan x +  =   tan x. Vijay thinks that

the solutions are x =  + n , x =  + n , x =

 + n , and x =  + n . Alicia thinks that the

solutions are x =  + n and x =  + n . Is

either of them correct? Explain your reasoning.

SOLUTION:  First, solve tan2 x – tan x +  =   tan x.

  On [0, 2π) tan x = 1 when x =  and x =  and 

tan x = when x =  and x = .

  Sample answer: Therefore, Vijay’s solutions are correct; however, they are not stated in the simplest

form. For example, his solutions of x =  + n and

x =  + n could simply be stated as x =  + 

n because when n = 1,  + nπ  is equivalent to

.

CHALLENGE  Solve each equation on the interval [0, 2π].

58. 16 sin5 x + 2 sin x = 12 sin3 x

SOLUTION:  

On [0, 2π) sin x = 0 when x = 0 and x = π, sin x =

when x =  and x = , sin x = when

x =  and x = , sin x =  – when x =  and 

x = , and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are 0, , , , , , , ,

, and .

59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3

SOLUTION:  

On [0, 2 ) sin x = 1 when x = , sin x  = –1 when

x = , sin x =  – when x =  and x = ,

and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are , , , , , and .

60. REASONING Are the solutions of csc x =  andcot2 x + 1 = 2 equivalent? If so, verify your answer algebraically. If not, explain your reasoning.

SOLUTION:  

 

 

The solutions to csc x =  are   and  . Using

a Pythagorean identity, cot2 x + 1 = 2 simplifies to csc2 x = 2 or csc x = ± . Therefore, there are

two additional solutions when csc x = – :  and

. So, the solutions of csc x =  and cot2 x + 1

= 2 are not equivalent.

OPEN ENDED Write a trigonometric equation that has each of the following solutions.

61. 

SOLUTION:  Sample answer: When sin x = 0, x = 0 and x = π.

When sin x = , x =  and x = . So, one

equation that has solutions of 0, π, , and  is 

= 0 or sin2 x –  = 0. This

can be rewritten as 2 sin2 x =  sin x.

62. 

SOLUTION:  

Sample answer: When cos x = , x =  and x =

. When cos x = – , x =  and x = .

So, one equation that has solutions of , , ,

and  is  = 0 or cos2 x

–  = 0. This can be rewritten as 4 cos2 x = 3.

63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying identities.

SOLUTION:  When solving an equation, you use properties of equality to manipulate each side of the equation to isolate a variable. For example:

  When verifying an identity, you transform an expression on one side of the identity into the expression on the other side through a series of algebraic steps.For example:

Verify = .  

Verify each identity.

64.  =

SOLUTION:  

65.  =

SOLUTION:  

66.  +  = 1

SOLUTION:  

+  

= cos2 θ + sin2 θ            Reciprocal Identities= 1                               Pythagorean Identity

Find the value of each expression using the given information.

67. tan ; sin θ = , tan  > 0

SOLUTION:  Use the Pythagorean Identity that involves sin θ.

Since tan  is positive and sin  is positive, cos θ

must be positive. So, .

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5-3 Solving Trigonometric Equations

Page 4: 5-3 Solving Trigonometric Equations the interval , the only solution is when x = 1.34. 39. x log x + 5x cos x = –2 SOLUTION: On the interval , the solutions are when x = 1.84 and

Solve each equation for all values of x.1. 5 sin x + 2 = sin x

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is  + 2nπ,  + 2nπ, .

2. 5 =  sec 2 x + 3

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

3. 2 = 4 cos2 x + 1

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on

the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

4. 4 tan x – 7 = 3 tan x – 6

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

5. 9 + cot2 x = 12

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

6. 2 – 10 sec x = 4 – 9 sec x

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

7. 3 csc x = 2 csc x +

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

8. 11 =  3 csc2 x + 7

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are , , , and . Solutions

on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

9. 6 tan2 x – 2 = 4

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

10. 9 + sin2 x = 10

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the interval

(– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the solutions is

 + 2nπ,  + 2nπ, .

11. 7 cot x –   = 4 cot x

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

12. 7 cos x = 5 cos x +

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ,   + 2nπ, .

Find all solutions of each equation on [0, 2 ).13. sin4 x + 2 sin2 x − 3 = 0

SOLUTION:  

On the interval [0, 2π),  when x =  and 

 when x = . Since  is not a real 

number, the equation  yields no additional solutions.

14. –2 sin x = –sin x cos x

SOLUTION:  

The equation cos x = 2 has no real solutions since the maximum value the cosine function can obtain is 1. On the interval [0, 2π), the equation  has solutions 0 and π.

15. 4 cot x = cot x sin2 x

SOLUTION:  

The equations sin x = 2 and sin x = –2 have no real solutions. On the interval [0, 2π), the equation cot x =

0 has solutions  and  .

16. csc2 x – csc x + 9 = 11

SOLUTION:  

On the interval [0, 2π), the equation csc x = –1 has a

solution of  and the equation csc x = 2 has

solutions of    and  .

17. cos3 x + cos2 x – cos x = 1

SOLUTION:  

On the interval [0, 2π), the equation cos x = 1 has a solution of 0 and the equation cos x = –1 has a solution of  π.

18. 2 sin2 x = sin x + 1

SOLUTION:  

On the interval [0, 2π), the equation sin x = 1 has a

solution of  and the equation sin x =  has  

solutions of    and  .

19. TENNIS A tennis ball leaves a racquet and heads toward a net 40 feet away. The height of the net is the same height as the initial height of the tennis ball. 

a. If the ball is hit at 50 feet per second, neglecting air resistance, use  to find the interval of possible angles of the ball needed to clear the net. b. Find  if the initial velocity remained the same but the distance to the net was 50 feet.

SOLUTION:  a.

The interval is [15.4°, 74.6°].   b.

If the distance to the net is 50 feet, then the angle would be 19.9° or 70.1°.

20. SKIING In the Olympics aerial skiing competition, skiers speed down a slope that launches them into the air, as shown. The maximum height a skier can

reach is given by hpeak =  , where g is

the acceleration due to gravity or 9.8 meters per second squared.

a. If a skier obtains a height of 5 meters above the end of the ramp, what was the skier’s initial speed? b. Use your answer from part a to determine how long it took the skier to reach the maximum height if tpeak = .

SOLUTION:  a.

The skier’s initial speed was 11.67 meters per second.   b.

  The time it took the skier to reach the maximum height was about 1.0 seconds.

Find all solutions of each equation on the interval [0, 2 ).

21. 1 = cot 2 x + csc x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

22. sec x = tan x + 1

SOLUTION:  

   

  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is 0.

23. tan 2 x = 1 – sec x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are 0,

, and .

24. csc x + cot x = 1

SOLUTION:  

 

x =  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is .

25. 2 – 2 cos2 x = sin x + 1

SOLUTION:  

Check

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

26. cos x – 4 = sin x – 4

SOLUTION:   

 

Therefore, on the interval [0, 2π) the only valid

solutions are and .

27. 3 sin x = 3 – 3 cos x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the only valid

solutions are and 0.

28. cot2 x csc2 x – cot2 x = 9

SOLUTION:  

is undefined. The solutions of  are 

  Check

 

 

 

On the interval , there are no real values of x for which , but  for 

29. sec2 x – 1 + tan x –  tan x =

SOLUTION:  

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

30. sec2 x tan2 x + 3 sec2 x – 2 tan2 x = 3

SOLUTION:  

 

The square of any real number must be greater than or equal to zero so sec2 x = –1 has no solutions. Therefore, on the interval [0, 2π) the only solutions are 0 and π.

31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive power PR of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical components, PH and PV.

Using the equations above, determine for what values of  PV and PH are equivalent.

SOLUTION:  

The sine and cosine have the same values in the

interval [0, 2π) at  and  .

Therefore, the components will be equivalent when

.

Find all solutions of each equation on the interval [0, 2 ).

32.  + cos x = 2

SOLUTION:  

On the interval [0, 2π), cos x =  when x =  and 

when x = .

33.  +  = –4

SOLUTION:  

On the interval [0, 2π),cos x = – when x =

 and when x = .

34.  +  = cos x

SOLUTION:  

On the interval [0, 2π) there are no values of x for

which sin x = –2 and the sin x = 1 only when x = .

However, tan  is undefined, so there are no 

solutions for the original equation.

35. cot x cos x + 1 =  + 

SOLUTION:  

On ,  when x =  and when x =

.

GRAPHING CALCULATOR Solve each equation on the interval [0, 2 ) by graphing. Round to the nearest hundredth.

36. 3 cos 2x = ex + 1

SOLUTION:  

On the interval , the only solution is when x = 0.33.

37. sin x + cos x = 3x

SOLUTION:  

On the interval , the only solution is when x = 0.41.

38. x2 = 2 cos x + x

SOLUTION:  

On the interval , the only solution is when x = 1.34.

39. x log x + 5x cos x = –2

SOLUTION:  

On the interval , the solutions are when x = 1.84 and when x = 4.49.

40. METEOROLOGY The average daily temperature in degrees Fahrenheit for a city can be modeled by t

= 8.05 cos + 66.95, where x is a function of

time, x = 1 represents January 15, x = 2 represents February 15, and so on. a. Use a graphing calculator to estimate the temperature on January 31. b. Approximate the number of months that the average daily temperature is greater than 70 throughout the entire month. c. Estimate the highest temperature of the year and the month in which it occurs.

SOLUTION:  a. Use a graphing calculator to graph t = 8.05

cos + 66.95. Because x = 1 represents

January 15, x = 1.5 represents January 31. Use the TRACE feature to determine the temperature whenx = 1.5.

Therefore, the temperature on January 31 is about 61.3º.   b. Graph y = 70. Then use the intersect feature under the CALC menu to determine the interval on

which 8.05 cos + 66.95 > 70.

So, 8.05 cos + 66.95 > 70 on the interval

3.74 < x < 8.26. Because x = 3.74 corresponds to about the middle of April, the temperature does not exceed 70º for the entire month until May. Therefore, the average daily temperature is greater than 70º for the entire month during the months of May, June, and July. Therefore, the average temperature was higher than 70º for three months.   c. Use the maximum feature under the CALC menu to find the maximum value of the function on [0, 12].

The maximum value is 75 when x ≈ 6. Therefore, the maximum average monthly temperature is 75º during the month of June.

Find the x-intercepts of each graph on the interval [0, 2 ).

41. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) cos x = 0 when x =  and x

= .

42. 

 

SOLUTION:  Let y = 0 and solve for x.

  On the interval [0, 2π) sin x = 0 when x = 0 and x = π.

43. 

SOLUTION:  Let y = 0 and solve for x.

 

On the interval [0, 2π) cot x = 1 when x =  and x

= .

44. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) tan x = 0 when x = 0 and x =

π, tan x = –1 when x =  and x = , and tan x

= 1 when x =  and x = .

Find all solutions of each equation on the interval [0, 4 ).

45. 4 tan x = 2 sec2 x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

46. 2 sin2 x + 1 = –3 sin x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, , , and .

47. csc x cot2 x = csc x

SOLUTION:   

On the interval [0, 4π) the solutions are , ,

, , , , , and .

48. sec x + 5 = 2 sec x + 3

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

49. GEOMETRY Consider the circle below.

a. The length s of  is given by s = r(2 ) where 0≤   ≤  . When s = 18 and AB = 14, the radius is r

= . Use a graphing calculator to find the

measure of 2 in radians. b. The area of the shaded region is given by A =

. Use a graphing calculator to find the

radian measure of θ  if the radius is 5 inches and the area is 36 square inches. Round to the nearest hundredth.

SOLUTION:  a. Rewrite the arclength formula using s = 18 and r

= .

On the graphing calculator, find the intersection of Y1 = 18sinθ and Y2 = 14θ.

The value of 2θ = 2(1.1968) or about 2.39 radians.     b. First, substitute into the given area formula and rearrange it.

  Using a graphing calculator, find the intersection of Y1 = 2.88 and Y2 = θ – sin θ.

When the area is 36 square inches and the radius is 5 inches, then the measure of θ is 3.01 radians.

Solve each inequality on the interval [0, 2 ).50. 1 > 2 sin x

SOLUTION:  Graph y = 1 and y = 2 sin x on [0, 2π). Use the intersect feature under the CALC menu to determine on what interval(s) 1 > 2 sin x.

  The graphs intersect at about 0.524 or  and about 

2.618 or . The graph of y = 1 is above the other

graph when x < and when x > . Therefore, 1

> 2 sin x on 0 ≤ x <    or   < x < 2π.

51. 0 < 2 cos x –

SOLUTION:  Graph y = 2 cos x –  on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0 < 2 cos x – .

  The zeros of y = 2 cos x –  are about 0.785 or  and about 5.498 or  . The graph of y = 0 is

below the other graph when 0 ≤ x <  or   < x <

2π. Therefore, 0 < 2 cos x –  on 0 ≤ x <  or   < 

x < 2π.

52.  ≥ 

SOLUTION:  

Graph y = and y =  on [0, 2π). Use

the intersect feature under the CALC menu to

determine on what interval(s) ≥  .

  The graphs intersect at about 1.047 or  and about 

2.094 or . The graph of y = is below the

other graph when   ≤ x ≤  .

Therefore, ≥   on  ≤ x ≤  .

53.  ≤ tan x cot x

SOLUTION:  

Graph y = and y = tan x cot x on [0, 2π).

Use the intersect feature under the CALC menu to

determine on what interval(s) ≤ tan x cot

x.  

  The graphs intersect at about 3.142 or π. The graph of y = tan x cot x is at or below the other graph when 0 ≤ x < 2π.  

Therefore, ≤ tan x cot x on 0 ≤ x < 2π.

54. cos x ≤

SOLUTION:  

Graph y = cos x and y =  on [0, 2π). Use the

intersect feature under the CALC menu to

determine on what interval(s) cos x ≤ .

  The graphs intersect at about 2.618 or  and about

3.665 or . The graph of y = cos x is below the

other graph when  ≤ x ≤  .

 

Therefore, cos x ≤ on  ≤ x ≤  .

55.  sin x − 1 < 0

SOLUTION:  Graph y =  sin x − 1. Use the zero feature underthe CALC menu to determine on what interval(s)

 sin x − 1 < 0.

  The zeros of  sin x − 1 are about 0.785 or  and

about 2.356 or . The graph is below the x-axis

when 0 ≤ x <  or   < x < 2π.

Therefore,  sin x − 1 < 0 on 0 ≤ x <  or   < 

x < 2π.

56. REFRACTION When light travels from one medium to another it bends or refracts, as shown.  

Refraction is described by n2 sinθ1 = n1 sinθ2, wheren1 is the index of refraction of the medium the light

is entering, n2 is the index of refraction of the medium the light is exiting, θ1 is the angle of

incidence, and θ2 is the angle of refraction. a. Find θ2 for each material shown if the angle of incidence is 40º and the index of refraction for air is 1.00.

b. If the angle of incidence is doubled to 80 , will the resulting angles of refraction be twice as large asthose found in part a?

SOLUTION:  a. glass:

  ice:

  plastic:

  water:

  b. Find the angle of refraction for one of the mediums using an angle of incidence of 80 . glass:

θ2 ≈ 25 when θ1 = 40 andθ2 ≈ 40.4 when θ1 = 80 . Therefore, the angle of refraction is not twice as large.

57. ERROR ANALYSIS Vijay and Alicia are solving tan2 x – tan x +  =   tan x. Vijay thinks that

the solutions are x =  + n , x =  + n , x =

 + n , and x =  + n . Alicia thinks that the

solutions are x =  + n and x =  + n . Is

either of them correct? Explain your reasoning.

SOLUTION:  First, solve tan2 x – tan x +  =   tan x.

  On [0, 2π) tan x = 1 when x =  and x =  and 

tan x = when x =  and x = .

  Sample answer: Therefore, Vijay’s solutions are correct; however, they are not stated in the simplest

form. For example, his solutions of x =  + n and

x =  + n could simply be stated as x =  + 

n because when n = 1,  + nπ  is equivalent to

.

CHALLENGE  Solve each equation on the interval [0, 2π].

58. 16 sin5 x + 2 sin x = 12 sin3 x

SOLUTION:  

On [0, 2π) sin x = 0 when x = 0 and x = π, sin x =

when x =  and x = , sin x = when

x =  and x = , sin x =  – when x =  and 

x = , and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are 0, , , , , , , ,

, and .

59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3

SOLUTION:  

On [0, 2 ) sin x = 1 when x = , sin x  = –1 when

x = , sin x =  – when x =  and x = ,

and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are , , , , , and .

60. REASONING Are the solutions of csc x =  andcot2 x + 1 = 2 equivalent? If so, verify your answer algebraically. If not, explain your reasoning.

SOLUTION:  

 

 

The solutions to csc x =  are   and  . Using

a Pythagorean identity, cot2 x + 1 = 2 simplifies to csc2 x = 2 or csc x = ± . Therefore, there are

two additional solutions when csc x = – :  and

. So, the solutions of csc x =  and cot2 x + 1

= 2 are not equivalent.

OPEN ENDED Write a trigonometric equation that has each of the following solutions.

61. 

SOLUTION:  Sample answer: When sin x = 0, x = 0 and x = π.

When sin x = , x =  and x = . So, one

equation that has solutions of 0, π, , and  is 

= 0 or sin2 x –  = 0. This

can be rewritten as 2 sin2 x =  sin x.

62. 

SOLUTION:  

Sample answer: When cos x = , x =  and x =

. When cos x = – , x =  and x = .

So, one equation that has solutions of , , ,

and  is  = 0 or cos2 x

–  = 0. This can be rewritten as 4 cos2 x = 3.

63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying identities.

SOLUTION:  When solving an equation, you use properties of equality to manipulate each side of the equation to isolate a variable. For example:

  When verifying an identity, you transform an expression on one side of the identity into the expression on the other side through a series of algebraic steps.For example:

Verify = .  

Verify each identity.

64.  =

SOLUTION:  

65.  =

SOLUTION:  

66.  +  = 1

SOLUTION:  

+  

= cos2 θ + sin2 θ            Reciprocal Identities= 1                               Pythagorean Identity

Find the value of each expression using the given information.

67. tan ; sin θ = , tan  > 0

SOLUTION:  Use the Pythagorean Identity that involves sin θ.

Since tan  is positive and sin  is positive, cos θ

must be positive. So, .

Solve each equation for all values of x.1. 5 sin x + 2 = sin x

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is  + 2nπ,  + 2nπ, .

2. 5 =  sec 2 x + 3

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

3. 2 = 4 cos2 x + 1

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on

the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

4. 4 tan x – 7 = 3 tan x – 6

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

5. 9 + cot2 x = 12

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

6. 2 – 10 sec x = 4 – 9 sec x

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

7. 3 csc x = 2 csc x +

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

8. 11 =  3 csc2 x + 7

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are , , , and . Solutions

on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

9. 6 tan2 x – 2 = 4

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

10. 9 + sin2 x = 10

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the interval

(– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the solutions is

 + 2nπ,  + 2nπ, .

11. 7 cot x –   = 4 cot x

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

12. 7 cos x = 5 cos x +

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ,   + 2nπ, .

Find all solutions of each equation on [0, 2 ).13. sin4 x + 2 sin2 x − 3 = 0

SOLUTION:  

On the interval [0, 2π),  when x =  and 

 when x = . Since  is not a real 

number, the equation  yields no additional solutions.

14. –2 sin x = –sin x cos x

SOLUTION:  

The equation cos x = 2 has no real solutions since the maximum value the cosine function can obtain is 1. On the interval [0, 2π), the equation  has solutions 0 and π.

15. 4 cot x = cot x sin2 x

SOLUTION:  

The equations sin x = 2 and sin x = –2 have no real solutions. On the interval [0, 2π), the equation cot x =

0 has solutions  and  .

16. csc2 x – csc x + 9 = 11

SOLUTION:  

On the interval [0, 2π), the equation csc x = –1 has a

solution of  and the equation csc x = 2 has

solutions of    and  .

17. cos3 x + cos2 x – cos x = 1

SOLUTION:  

On the interval [0, 2π), the equation cos x = 1 has a solution of 0 and the equation cos x = –1 has a solution of  π.

18. 2 sin2 x = sin x + 1

SOLUTION:  

On the interval [0, 2π), the equation sin x = 1 has a

solution of  and the equation sin x =  has  

solutions of    and  .

19. TENNIS A tennis ball leaves a racquet and heads toward a net 40 feet away. The height of the net is the same height as the initial height of the tennis ball. 

a. If the ball is hit at 50 feet per second, neglecting air resistance, use  to find the interval of possible angles of the ball needed to clear the net. b. Find  if the initial velocity remained the same but the distance to the net was 50 feet.

SOLUTION:  a.

The interval is [15.4°, 74.6°].   b.

If the distance to the net is 50 feet, then the angle would be 19.9° or 70.1°.

20. SKIING In the Olympics aerial skiing competition, skiers speed down a slope that launches them into the air, as shown. The maximum height a skier can

reach is given by hpeak =  , where g is

the acceleration due to gravity or 9.8 meters per second squared.

a. If a skier obtains a height of 5 meters above the end of the ramp, what was the skier’s initial speed? b. Use your answer from part a to determine how long it took the skier to reach the maximum height if tpeak = .

SOLUTION:  a.

The skier’s initial speed was 11.67 meters per second.   b.

  The time it took the skier to reach the maximum height was about 1.0 seconds.

Find all solutions of each equation on the interval [0, 2 ).

21. 1 = cot 2 x + csc x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

22. sec x = tan x + 1

SOLUTION:  

   

  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is 0.

23. tan 2 x = 1 – sec x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are 0,

, and .

24. csc x + cot x = 1

SOLUTION:  

 

x =  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is .

25. 2 – 2 cos2 x = sin x + 1

SOLUTION:  

Check

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

26. cos x – 4 = sin x – 4

SOLUTION:   

 

Therefore, on the interval [0, 2π) the only valid

solutions are and .

27. 3 sin x = 3 – 3 cos x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the only valid

solutions are and 0.

28. cot2 x csc2 x – cot2 x = 9

SOLUTION:  

is undefined. The solutions of  are 

  Check

 

 

 

On the interval , there are no real values of x for which , but  for 

29. sec2 x – 1 + tan x –  tan x =

SOLUTION:  

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

30. sec2 x tan2 x + 3 sec2 x – 2 tan2 x = 3

SOLUTION:  

 

The square of any real number must be greater than or equal to zero so sec2 x = –1 has no solutions. Therefore, on the interval [0, 2π) the only solutions are 0 and π.

31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive power PR of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical components, PH and PV.

Using the equations above, determine for what values of  PV and PH are equivalent.

SOLUTION:  

The sine and cosine have the same values in the

interval [0, 2π) at  and  .

Therefore, the components will be equivalent when

.

Find all solutions of each equation on the interval [0, 2 ).

32.  + cos x = 2

SOLUTION:  

On the interval [0, 2π), cos x =  when x =  and 

when x = .

33.  +  = –4

SOLUTION:  

On the interval [0, 2π),cos x = – when x =

 and when x = .

34.  +  = cos x

SOLUTION:  

On the interval [0, 2π) there are no values of x for

which sin x = –2 and the sin x = 1 only when x = .

However, tan  is undefined, so there are no 

solutions for the original equation.

35. cot x cos x + 1 =  + 

SOLUTION:  

On ,  when x =  and when x =

.

GRAPHING CALCULATOR Solve each equation on the interval [0, 2 ) by graphing. Round to the nearest hundredth.

36. 3 cos 2x = ex + 1

SOLUTION:  

On the interval , the only solution is when x = 0.33.

37. sin x + cos x = 3x

SOLUTION:  

On the interval , the only solution is when x = 0.41.

38. x2 = 2 cos x + x

SOLUTION:  

On the interval , the only solution is when x = 1.34.

39. x log x + 5x cos x = –2

SOLUTION:  

On the interval , the solutions are when x = 1.84 and when x = 4.49.

40. METEOROLOGY The average daily temperature in degrees Fahrenheit for a city can be modeled by t

= 8.05 cos + 66.95, where x is a function of

time, x = 1 represents January 15, x = 2 represents February 15, and so on. a. Use a graphing calculator to estimate the temperature on January 31. b. Approximate the number of months that the average daily temperature is greater than 70 throughout the entire month. c. Estimate the highest temperature of the year and the month in which it occurs.

SOLUTION:  a. Use a graphing calculator to graph t = 8.05

cos + 66.95. Because x = 1 represents

January 15, x = 1.5 represents January 31. Use the TRACE feature to determine the temperature whenx = 1.5.

Therefore, the temperature on January 31 is about 61.3º.   b. Graph y = 70. Then use the intersect feature under the CALC menu to determine the interval on

which 8.05 cos + 66.95 > 70.

So, 8.05 cos + 66.95 > 70 on the interval

3.74 < x < 8.26. Because x = 3.74 corresponds to about the middle of April, the temperature does not exceed 70º for the entire month until May. Therefore, the average daily temperature is greater than 70º for the entire month during the months of May, June, and July. Therefore, the average temperature was higher than 70º for three months.   c. Use the maximum feature under the CALC menu to find the maximum value of the function on [0, 12].

The maximum value is 75 when x ≈ 6. Therefore, the maximum average monthly temperature is 75º during the month of June.

Find the x-intercepts of each graph on the interval [0, 2 ).

41. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) cos x = 0 when x =  and x

= .

42. 

 

SOLUTION:  Let y = 0 and solve for x.

  On the interval [0, 2π) sin x = 0 when x = 0 and x = π.

43. 

SOLUTION:  Let y = 0 and solve for x.

 

On the interval [0, 2π) cot x = 1 when x =  and x

= .

44. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) tan x = 0 when x = 0 and x =

π, tan x = –1 when x =  and x = , and tan x

= 1 when x =  and x = .

Find all solutions of each equation on the interval [0, 4 ).

45. 4 tan x = 2 sec2 x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

46. 2 sin2 x + 1 = –3 sin x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, , , and .

47. csc x cot2 x = csc x

SOLUTION:   

On the interval [0, 4π) the solutions are , ,

, , , , , and .

48. sec x + 5 = 2 sec x + 3

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

49. GEOMETRY Consider the circle below.

a. The length s of  is given by s = r(2 ) where 0≤   ≤  . When s = 18 and AB = 14, the radius is r

= . Use a graphing calculator to find the

measure of 2 in radians. b. The area of the shaded region is given by A =

. Use a graphing calculator to find the

radian measure of θ  if the radius is 5 inches and the area is 36 square inches. Round to the nearest hundredth.

SOLUTION:  a. Rewrite the arclength formula using s = 18 and r

= .

On the graphing calculator, find the intersection of Y1 = 18sinθ and Y2 = 14θ.

The value of 2θ = 2(1.1968) or about 2.39 radians.     b. First, substitute into the given area formula and rearrange it.

  Using a graphing calculator, find the intersection of Y1 = 2.88 and Y2 = θ – sin θ.

When the area is 36 square inches and the radius is 5 inches, then the measure of θ is 3.01 radians.

Solve each inequality on the interval [0, 2 ).50. 1 > 2 sin x

SOLUTION:  Graph y = 1 and y = 2 sin x on [0, 2π). Use the intersect feature under the CALC menu to determine on what interval(s) 1 > 2 sin x.

  The graphs intersect at about 0.524 or  and about 

2.618 or . The graph of y = 1 is above the other

graph when x < and when x > . Therefore, 1

> 2 sin x on 0 ≤ x <    or   < x < 2π.

51. 0 < 2 cos x –

SOLUTION:  Graph y = 2 cos x –  on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0 < 2 cos x – .

  The zeros of y = 2 cos x –  are about 0.785 or  and about 5.498 or  . The graph of y = 0 is

below the other graph when 0 ≤ x <  or   < x <

2π. Therefore, 0 < 2 cos x –  on 0 ≤ x <  or   < 

x < 2π.

52.  ≥ 

SOLUTION:  

Graph y = and y =  on [0, 2π). Use

the intersect feature under the CALC menu to

determine on what interval(s) ≥  .

  The graphs intersect at about 1.047 or  and about 

2.094 or . The graph of y = is below the

other graph when   ≤ x ≤  .

Therefore, ≥   on  ≤ x ≤  .

53.  ≤ tan x cot x

SOLUTION:  

Graph y = and y = tan x cot x on [0, 2π).

Use the intersect feature under the CALC menu to

determine on what interval(s) ≤ tan x cot

x.  

  The graphs intersect at about 3.142 or π. The graph of y = tan x cot x is at or below the other graph when 0 ≤ x < 2π.  

Therefore, ≤ tan x cot x on 0 ≤ x < 2π.

54. cos x ≤

SOLUTION:  

Graph y = cos x and y =  on [0, 2π). Use the

intersect feature under the CALC menu to

determine on what interval(s) cos x ≤ .

  The graphs intersect at about 2.618 or  and about

3.665 or . The graph of y = cos x is below the

other graph when  ≤ x ≤  .

 

Therefore, cos x ≤ on  ≤ x ≤  .

55.  sin x − 1 < 0

SOLUTION:  Graph y =  sin x − 1. Use the zero feature underthe CALC menu to determine on what interval(s)

 sin x − 1 < 0.

  The zeros of  sin x − 1 are about 0.785 or  and

about 2.356 or . The graph is below the x-axis

when 0 ≤ x <  or   < x < 2π.

Therefore,  sin x − 1 < 0 on 0 ≤ x <  or   < 

x < 2π.

56. REFRACTION When light travels from one medium to another it bends or refracts, as shown.  

Refraction is described by n2 sinθ1 = n1 sinθ2, wheren1 is the index of refraction of the medium the light

is entering, n2 is the index of refraction of the medium the light is exiting, θ1 is the angle of

incidence, and θ2 is the angle of refraction. a. Find θ2 for each material shown if the angle of incidence is 40º and the index of refraction for air is 1.00.

b. If the angle of incidence is doubled to 80 , will the resulting angles of refraction be twice as large asthose found in part a?

SOLUTION:  a. glass:

  ice:

  plastic:

  water:

  b. Find the angle of refraction for one of the mediums using an angle of incidence of 80 . glass:

θ2 ≈ 25 when θ1 = 40 andθ2 ≈ 40.4 when θ1 = 80 . Therefore, the angle of refraction is not twice as large.

57. ERROR ANALYSIS Vijay and Alicia are solving tan2 x – tan x +  =   tan x. Vijay thinks that

the solutions are x =  + n , x =  + n , x =

 + n , and x =  + n . Alicia thinks that the

solutions are x =  + n and x =  + n . Is

either of them correct? Explain your reasoning.

SOLUTION:  First, solve tan2 x – tan x +  =   tan x.

  On [0, 2π) tan x = 1 when x =  and x =  and 

tan x = when x =  and x = .

  Sample answer: Therefore, Vijay’s solutions are correct; however, they are not stated in the simplest

form. For example, his solutions of x =  + n and

x =  + n could simply be stated as x =  + 

n because when n = 1,  + nπ  is equivalent to

.

CHALLENGE  Solve each equation on the interval [0, 2π].

58. 16 sin5 x + 2 sin x = 12 sin3 x

SOLUTION:  

On [0, 2π) sin x = 0 when x = 0 and x = π, sin x =

when x =  and x = , sin x = when

x =  and x = , sin x =  – when x =  and 

x = , and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are 0, , , , , , , ,

, and .

59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3

SOLUTION:  

On [0, 2 ) sin x = 1 when x = , sin x  = –1 when

x = , sin x =  – when x =  and x = ,

and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are , , , , , and .

60. REASONING Are the solutions of csc x =  andcot2 x + 1 = 2 equivalent? If so, verify your answer algebraically. If not, explain your reasoning.

SOLUTION:  

 

 

The solutions to csc x =  are   and  . Using

a Pythagorean identity, cot2 x + 1 = 2 simplifies to csc2 x = 2 or csc x = ± . Therefore, there are

two additional solutions when csc x = – :  and

. So, the solutions of csc x =  and cot2 x + 1

= 2 are not equivalent.

OPEN ENDED Write a trigonometric equation that has each of the following solutions.

61. 

SOLUTION:  Sample answer: When sin x = 0, x = 0 and x = π.

When sin x = , x =  and x = . So, one

equation that has solutions of 0, π, , and  is 

= 0 or sin2 x –  = 0. This

can be rewritten as 2 sin2 x =  sin x.

62. 

SOLUTION:  

Sample answer: When cos x = , x =  and x =

. When cos x = – , x =  and x = .

So, one equation that has solutions of , , ,

and  is  = 0 or cos2 x

–  = 0. This can be rewritten as 4 cos2 x = 3.

63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying identities.

SOLUTION:  When solving an equation, you use properties of equality to manipulate each side of the equation to isolate a variable. For example:

  When verifying an identity, you transform an expression on one side of the identity into the expression on the other side through a series of algebraic steps.For example:

Verify = .  

Verify each identity.

64.  =

SOLUTION:  

65.  =

SOLUTION:  

66.  +  = 1

SOLUTION:  

+  

= cos2 θ + sin2 θ            Reciprocal Identities= 1                               Pythagorean Identity

Find the value of each expression using the given information.

67. tan ; sin θ = , tan  > 0

SOLUTION:  Use the Pythagorean Identity that involves sin θ.

Since tan  is positive and sin  is positive, cos θ

must be positive. So, .

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5-3 Solving Trigonometric Equations

Page 5: 5-3 Solving Trigonometric Equations the interval , the only solution is when x = 1.34. 39. x log x + 5x cos x = –2 SOLUTION: On the interval , the solutions are when x = 1.84 and

Solve each equation for all values of x.1. 5 sin x + 2 = sin x

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is  + 2nπ,  + 2nπ, .

2. 5 =  sec 2 x + 3

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

3. 2 = 4 cos2 x + 1

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on

the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

4. 4 tan x – 7 = 3 tan x – 6

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

5. 9 + cot2 x = 12

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

6. 2 – 10 sec x = 4 – 9 sec x

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

7. 3 csc x = 2 csc x +

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

8. 11 =  3 csc2 x + 7

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are , , , and . Solutions

on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

9. 6 tan2 x – 2 = 4

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

10. 9 + sin2 x = 10

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the interval

(– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the solutions is

 + 2nπ,  + 2nπ, .

11. 7 cot x –   = 4 cot x

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

12. 7 cos x = 5 cos x +

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ,   + 2nπ, .

Find all solutions of each equation on [0, 2 ).13. sin4 x + 2 sin2 x − 3 = 0

SOLUTION:  

On the interval [0, 2π),  when x =  and 

 when x = . Since  is not a real 

number, the equation  yields no additional solutions.

14. –2 sin x = –sin x cos x

SOLUTION:  

The equation cos x = 2 has no real solutions since the maximum value the cosine function can obtain is 1. On the interval [0, 2π), the equation  has solutions 0 and π.

15. 4 cot x = cot x sin2 x

SOLUTION:  

The equations sin x = 2 and sin x = –2 have no real solutions. On the interval [0, 2π), the equation cot x =

0 has solutions  and  .

16. csc2 x – csc x + 9 = 11

SOLUTION:  

On the interval [0, 2π), the equation csc x = –1 has a

solution of  and the equation csc x = 2 has

solutions of    and  .

17. cos3 x + cos2 x – cos x = 1

SOLUTION:  

On the interval [0, 2π), the equation cos x = 1 has a solution of 0 and the equation cos x = –1 has a solution of  π.

18. 2 sin2 x = sin x + 1

SOLUTION:  

On the interval [0, 2π), the equation sin x = 1 has a

solution of  and the equation sin x =  has  

solutions of    and  .

19. TENNIS A tennis ball leaves a racquet and heads toward a net 40 feet away. The height of the net is the same height as the initial height of the tennis ball. 

a. If the ball is hit at 50 feet per second, neglecting air resistance, use  to find the interval of possible angles of the ball needed to clear the net. b. Find  if the initial velocity remained the same but the distance to the net was 50 feet.

SOLUTION:  a.

The interval is [15.4°, 74.6°].   b.

If the distance to the net is 50 feet, then the angle would be 19.9° or 70.1°.

20. SKIING In the Olympics aerial skiing competition, skiers speed down a slope that launches them into the air, as shown. The maximum height a skier can

reach is given by hpeak =  , where g is

the acceleration due to gravity or 9.8 meters per second squared.

a. If a skier obtains a height of 5 meters above the end of the ramp, what was the skier’s initial speed? b. Use your answer from part a to determine how long it took the skier to reach the maximum height if tpeak = .

SOLUTION:  a.

The skier’s initial speed was 11.67 meters per second.   b.

  The time it took the skier to reach the maximum height was about 1.0 seconds.

Find all solutions of each equation on the interval [0, 2 ).

21. 1 = cot 2 x + csc x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

22. sec x = tan x + 1

SOLUTION:  

   

  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is 0.

23. tan 2 x = 1 – sec x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are 0,

, and .

24. csc x + cot x = 1

SOLUTION:  

 

x =  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is .

25. 2 – 2 cos2 x = sin x + 1

SOLUTION:  

Check

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

26. cos x – 4 = sin x – 4

SOLUTION:   

 

Therefore, on the interval [0, 2π) the only valid

solutions are and .

27. 3 sin x = 3 – 3 cos x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the only valid

solutions are and 0.

28. cot2 x csc2 x – cot2 x = 9

SOLUTION:  

is undefined. The solutions of  are 

  Check

 

 

 

On the interval , there are no real values of x for which , but  for 

29. sec2 x – 1 + tan x –  tan x =

SOLUTION:  

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

30. sec2 x tan2 x + 3 sec2 x – 2 tan2 x = 3

SOLUTION:  

 

The square of any real number must be greater than or equal to zero so sec2 x = –1 has no solutions. Therefore, on the interval [0, 2π) the only solutions are 0 and π.

31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive power PR of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical components, PH and PV.

Using the equations above, determine for what values of  PV and PH are equivalent.

SOLUTION:  

The sine and cosine have the same values in the

interval [0, 2π) at  and  .

Therefore, the components will be equivalent when

.

Find all solutions of each equation on the interval [0, 2 ).

32.  + cos x = 2

SOLUTION:  

On the interval [0, 2π), cos x =  when x =  and 

when x = .

33.  +  = –4

SOLUTION:  

On the interval [0, 2π),cos x = – when x =

 and when x = .

34.  +  = cos x

SOLUTION:  

On the interval [0, 2π) there are no values of x for

which sin x = –2 and the sin x = 1 only when x = .

However, tan  is undefined, so there are no 

solutions for the original equation.

35. cot x cos x + 1 =  + 

SOLUTION:  

On ,  when x =  and when x =

.

GRAPHING CALCULATOR Solve each equation on the interval [0, 2 ) by graphing. Round to the nearest hundredth.

36. 3 cos 2x = ex + 1

SOLUTION:  

On the interval , the only solution is when x = 0.33.

37. sin x + cos x = 3x

SOLUTION:  

On the interval , the only solution is when x = 0.41.

38. x2 = 2 cos x + x

SOLUTION:  

On the interval , the only solution is when x = 1.34.

39. x log x + 5x cos x = –2

SOLUTION:  

On the interval , the solutions are when x = 1.84 and when x = 4.49.

40. METEOROLOGY The average daily temperature in degrees Fahrenheit for a city can be modeled by t

= 8.05 cos + 66.95, where x is a function of

time, x = 1 represents January 15, x = 2 represents February 15, and so on. a. Use a graphing calculator to estimate the temperature on January 31. b. Approximate the number of months that the average daily temperature is greater than 70 throughout the entire month. c. Estimate the highest temperature of the year and the month in which it occurs.

SOLUTION:  a. Use a graphing calculator to graph t = 8.05

cos + 66.95. Because x = 1 represents

January 15, x = 1.5 represents January 31. Use the TRACE feature to determine the temperature whenx = 1.5.

Therefore, the temperature on January 31 is about 61.3º.   b. Graph y = 70. Then use the intersect feature under the CALC menu to determine the interval on

which 8.05 cos + 66.95 > 70.

So, 8.05 cos + 66.95 > 70 on the interval

3.74 < x < 8.26. Because x = 3.74 corresponds to about the middle of April, the temperature does not exceed 70º for the entire month until May. Therefore, the average daily temperature is greater than 70º for the entire month during the months of May, June, and July. Therefore, the average temperature was higher than 70º for three months.   c. Use the maximum feature under the CALC menu to find the maximum value of the function on [0, 12].

The maximum value is 75 when x ≈ 6. Therefore, the maximum average monthly temperature is 75º during the month of June.

Find the x-intercepts of each graph on the interval [0, 2 ).

41. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) cos x = 0 when x =  and x

= .

42. 

 

SOLUTION:  Let y = 0 and solve for x.

  On the interval [0, 2π) sin x = 0 when x = 0 and x = π.

43. 

SOLUTION:  Let y = 0 and solve for x.

 

On the interval [0, 2π) cot x = 1 when x =  and x

= .

44. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) tan x = 0 when x = 0 and x =

π, tan x = –1 when x =  and x = , and tan x

= 1 when x =  and x = .

Find all solutions of each equation on the interval [0, 4 ).

45. 4 tan x = 2 sec2 x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

46. 2 sin2 x + 1 = –3 sin x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, , , and .

47. csc x cot2 x = csc x

SOLUTION:   

On the interval [0, 4π) the solutions are , ,

, , , , , and .

48. sec x + 5 = 2 sec x + 3

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

49. GEOMETRY Consider the circle below.

a. The length s of  is given by s = r(2 ) where 0≤   ≤  . When s = 18 and AB = 14, the radius is r

= . Use a graphing calculator to find the

measure of 2 in radians. b. The area of the shaded region is given by A =

. Use a graphing calculator to find the

radian measure of θ  if the radius is 5 inches and the area is 36 square inches. Round to the nearest hundredth.

SOLUTION:  a. Rewrite the arclength formula using s = 18 and r

= .

On the graphing calculator, find the intersection of Y1 = 18sinθ and Y2 = 14θ.

The value of 2θ = 2(1.1968) or about 2.39 radians.     b. First, substitute into the given area formula and rearrange it.

  Using a graphing calculator, find the intersection of Y1 = 2.88 and Y2 = θ – sin θ.

When the area is 36 square inches and the radius is 5 inches, then the measure of θ is 3.01 radians.

Solve each inequality on the interval [0, 2 ).50. 1 > 2 sin x

SOLUTION:  Graph y = 1 and y = 2 sin x on [0, 2π). Use the intersect feature under the CALC menu to determine on what interval(s) 1 > 2 sin x.

  The graphs intersect at about 0.524 or  and about 

2.618 or . The graph of y = 1 is above the other

graph when x < and when x > . Therefore, 1

> 2 sin x on 0 ≤ x <    or   < x < 2π.

51. 0 < 2 cos x –

SOLUTION:  Graph y = 2 cos x –  on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0 < 2 cos x – .

  The zeros of y = 2 cos x –  are about 0.785 or  and about 5.498 or  . The graph of y = 0 is

below the other graph when 0 ≤ x <  or   < x <

2π. Therefore, 0 < 2 cos x –  on 0 ≤ x <  or   < 

x < 2π.

52.  ≥ 

SOLUTION:  

Graph y = and y =  on [0, 2π). Use

the intersect feature under the CALC menu to

determine on what interval(s) ≥  .

  The graphs intersect at about 1.047 or  and about 

2.094 or . The graph of y = is below the

other graph when   ≤ x ≤  .

Therefore, ≥   on  ≤ x ≤  .

53.  ≤ tan x cot x

SOLUTION:  

Graph y = and y = tan x cot x on [0, 2π).

Use the intersect feature under the CALC menu to

determine on what interval(s) ≤ tan x cot

x.  

  The graphs intersect at about 3.142 or π. The graph of y = tan x cot x is at or below the other graph when 0 ≤ x < 2π.  

Therefore, ≤ tan x cot x on 0 ≤ x < 2π.

54. cos x ≤

SOLUTION:  

Graph y = cos x and y =  on [0, 2π). Use the

intersect feature under the CALC menu to

determine on what interval(s) cos x ≤ .

  The graphs intersect at about 2.618 or  and about

3.665 or . The graph of y = cos x is below the

other graph when  ≤ x ≤  .

 

Therefore, cos x ≤ on  ≤ x ≤  .

55.  sin x − 1 < 0

SOLUTION:  Graph y =  sin x − 1. Use the zero feature underthe CALC menu to determine on what interval(s)

 sin x − 1 < 0.

  The zeros of  sin x − 1 are about 0.785 or  and

about 2.356 or . The graph is below the x-axis

when 0 ≤ x <  or   < x < 2π.

Therefore,  sin x − 1 < 0 on 0 ≤ x <  or   < 

x < 2π.

56. REFRACTION When light travels from one medium to another it bends or refracts, as shown.  

Refraction is described by n2 sinθ1 = n1 sinθ2, wheren1 is the index of refraction of the medium the light

is entering, n2 is the index of refraction of the medium the light is exiting, θ1 is the angle of

incidence, and θ2 is the angle of refraction. a. Find θ2 for each material shown if the angle of incidence is 40º and the index of refraction for air is 1.00.

b. If the angle of incidence is doubled to 80 , will the resulting angles of refraction be twice as large asthose found in part a?

SOLUTION:  a. glass:

  ice:

  plastic:

  water:

  b. Find the angle of refraction for one of the mediums using an angle of incidence of 80 . glass:

θ2 ≈ 25 when θ1 = 40 andθ2 ≈ 40.4 when θ1 = 80 . Therefore, the angle of refraction is not twice as large.

57. ERROR ANALYSIS Vijay and Alicia are solving tan2 x – tan x +  =   tan x. Vijay thinks that

the solutions are x =  + n , x =  + n , x =

 + n , and x =  + n . Alicia thinks that the

solutions are x =  + n and x =  + n . Is

either of them correct? Explain your reasoning.

SOLUTION:  First, solve tan2 x – tan x +  =   tan x.

  On [0, 2π) tan x = 1 when x =  and x =  and 

tan x = when x =  and x = .

  Sample answer: Therefore, Vijay’s solutions are correct; however, they are not stated in the simplest

form. For example, his solutions of x =  + n and

x =  + n could simply be stated as x =  + 

n because when n = 1,  + nπ  is equivalent to

.

CHALLENGE  Solve each equation on the interval [0, 2π].

58. 16 sin5 x + 2 sin x = 12 sin3 x

SOLUTION:  

On [0, 2π) sin x = 0 when x = 0 and x = π, sin x =

when x =  and x = , sin x = when

x =  and x = , sin x =  – when x =  and 

x = , and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are 0, , , , , , , ,

, and .

59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3

SOLUTION:  

On [0, 2 ) sin x = 1 when x = , sin x  = –1 when

x = , sin x =  – when x =  and x = ,

and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are , , , , , and .

60. REASONING Are the solutions of csc x =  andcot2 x + 1 = 2 equivalent? If so, verify your answer algebraically. If not, explain your reasoning.

SOLUTION:  

 

 

The solutions to csc x =  are   and  . Using

a Pythagorean identity, cot2 x + 1 = 2 simplifies to csc2 x = 2 or csc x = ± . Therefore, there are

two additional solutions when csc x = – :  and

. So, the solutions of csc x =  and cot2 x + 1

= 2 are not equivalent.

OPEN ENDED Write a trigonometric equation that has each of the following solutions.

61. 

SOLUTION:  Sample answer: When sin x = 0, x = 0 and x = π.

When sin x = , x =  and x = . So, one

equation that has solutions of 0, π, , and  is 

= 0 or sin2 x –  = 0. This

can be rewritten as 2 sin2 x =  sin x.

62. 

SOLUTION:  

Sample answer: When cos x = , x =  and x =

. When cos x = – , x =  and x = .

So, one equation that has solutions of , , ,

and  is  = 0 or cos2 x

–  = 0. This can be rewritten as 4 cos2 x = 3.

63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying identities.

SOLUTION:  When solving an equation, you use properties of equality to manipulate each side of the equation to isolate a variable. For example:

  When verifying an identity, you transform an expression on one side of the identity into the expression on the other side through a series of algebraic steps.For example:

Verify = .  

Verify each identity.

64.  =

SOLUTION:  

65.  =

SOLUTION:  

66.  +  = 1

SOLUTION:  

+  

= cos2 θ + sin2 θ            Reciprocal Identities= 1                               Pythagorean Identity

Find the value of each expression using the given information.

67. tan ; sin θ = , tan  > 0

SOLUTION:  Use the Pythagorean Identity that involves sin θ.

Since tan  is positive and sin  is positive, cos θ

must be positive. So, .

Solve each equation for all values of x.1. 5 sin x + 2 = sin x

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is  + 2nπ,  + 2nπ, .

2. 5 =  sec 2 x + 3

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

3. 2 = 4 cos2 x + 1

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on

the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

4. 4 tan x – 7 = 3 tan x – 6

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

5. 9 + cot2 x = 12

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

6. 2 – 10 sec x = 4 – 9 sec x

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

7. 3 csc x = 2 csc x +

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

8. 11 =  3 csc2 x + 7

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are , , , and . Solutions

on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

9. 6 tan2 x – 2 = 4

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

10. 9 + sin2 x = 10

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the interval

(– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the solutions is

 + 2nπ,  + 2nπ, .

11. 7 cot x –   = 4 cot x

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

12. 7 cos x = 5 cos x +

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ,   + 2nπ, .

Find all solutions of each equation on [0, 2 ).13. sin4 x + 2 sin2 x − 3 = 0

SOLUTION:  

On the interval [0, 2π),  when x =  and 

 when x = . Since  is not a real 

number, the equation  yields no additional solutions.

14. –2 sin x = –sin x cos x

SOLUTION:  

The equation cos x = 2 has no real solutions since the maximum value the cosine function can obtain is 1. On the interval [0, 2π), the equation  has solutions 0 and π.

15. 4 cot x = cot x sin2 x

SOLUTION:  

The equations sin x = 2 and sin x = –2 have no real solutions. On the interval [0, 2π), the equation cot x =

0 has solutions  and  .

16. csc2 x – csc x + 9 = 11

SOLUTION:  

On the interval [0, 2π), the equation csc x = –1 has a

solution of  and the equation csc x = 2 has

solutions of    and  .

17. cos3 x + cos2 x – cos x = 1

SOLUTION:  

On the interval [0, 2π), the equation cos x = 1 has a solution of 0 and the equation cos x = –1 has a solution of  π.

18. 2 sin2 x = sin x + 1

SOLUTION:  

On the interval [0, 2π), the equation sin x = 1 has a

solution of  and the equation sin x =  has  

solutions of    and  .

19. TENNIS A tennis ball leaves a racquet and heads toward a net 40 feet away. The height of the net is the same height as the initial height of the tennis ball. 

a. If the ball is hit at 50 feet per second, neglecting air resistance, use  to find the interval of possible angles of the ball needed to clear the net. b. Find  if the initial velocity remained the same but the distance to the net was 50 feet.

SOLUTION:  a.

The interval is [15.4°, 74.6°].   b.

If the distance to the net is 50 feet, then the angle would be 19.9° or 70.1°.

20. SKIING In the Olympics aerial skiing competition, skiers speed down a slope that launches them into the air, as shown. The maximum height a skier can

reach is given by hpeak =  , where g is

the acceleration due to gravity or 9.8 meters per second squared.

a. If a skier obtains a height of 5 meters above the end of the ramp, what was the skier’s initial speed? b. Use your answer from part a to determine how long it took the skier to reach the maximum height if tpeak = .

SOLUTION:  a.

The skier’s initial speed was 11.67 meters per second.   b.

  The time it took the skier to reach the maximum height was about 1.0 seconds.

Find all solutions of each equation on the interval [0, 2 ).

21. 1 = cot 2 x + csc x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

22. sec x = tan x + 1

SOLUTION:  

   

  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is 0.

23. tan 2 x = 1 – sec x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are 0,

, and .

24. csc x + cot x = 1

SOLUTION:  

 

x =  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is .

25. 2 – 2 cos2 x = sin x + 1

SOLUTION:  

Check

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

26. cos x – 4 = sin x – 4

SOLUTION:   

 

Therefore, on the interval [0, 2π) the only valid

solutions are and .

27. 3 sin x = 3 – 3 cos x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the only valid

solutions are and 0.

28. cot2 x csc2 x – cot2 x = 9

SOLUTION:  

is undefined. The solutions of  are 

  Check

 

 

 

On the interval , there are no real values of x for which , but  for 

29. sec2 x – 1 + tan x –  tan x =

SOLUTION:  

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

30. sec2 x tan2 x + 3 sec2 x – 2 tan2 x = 3

SOLUTION:  

 

The square of any real number must be greater than or equal to zero so sec2 x = –1 has no solutions. Therefore, on the interval [0, 2π) the only solutions are 0 and π.

31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive power PR of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical components, PH and PV.

Using the equations above, determine for what values of  PV and PH are equivalent.

SOLUTION:  

The sine and cosine have the same values in the

interval [0, 2π) at  and  .

Therefore, the components will be equivalent when

.

Find all solutions of each equation on the interval [0, 2 ).

32.  + cos x = 2

SOLUTION:  

On the interval [0, 2π), cos x =  when x =  and 

when x = .

33.  +  = –4

SOLUTION:  

On the interval [0, 2π),cos x = – when x =

 and when x = .

34.  +  = cos x

SOLUTION:  

On the interval [0, 2π) there are no values of x for

which sin x = –2 and the sin x = 1 only when x = .

However, tan  is undefined, so there are no 

solutions for the original equation.

35. cot x cos x + 1 =  + 

SOLUTION:  

On ,  when x =  and when x =

.

GRAPHING CALCULATOR Solve each equation on the interval [0, 2 ) by graphing. Round to the nearest hundredth.

36. 3 cos 2x = ex + 1

SOLUTION:  

On the interval , the only solution is when x = 0.33.

37. sin x + cos x = 3x

SOLUTION:  

On the interval , the only solution is when x = 0.41.

38. x2 = 2 cos x + x

SOLUTION:  

On the interval , the only solution is when x = 1.34.

39. x log x + 5x cos x = –2

SOLUTION:  

On the interval , the solutions are when x = 1.84 and when x = 4.49.

40. METEOROLOGY The average daily temperature in degrees Fahrenheit for a city can be modeled by t

= 8.05 cos + 66.95, where x is a function of

time, x = 1 represents January 15, x = 2 represents February 15, and so on. a. Use a graphing calculator to estimate the temperature on January 31. b. Approximate the number of months that the average daily temperature is greater than 70 throughout the entire month. c. Estimate the highest temperature of the year and the month in which it occurs.

SOLUTION:  a. Use a graphing calculator to graph t = 8.05

cos + 66.95. Because x = 1 represents

January 15, x = 1.5 represents January 31. Use the TRACE feature to determine the temperature whenx = 1.5.

Therefore, the temperature on January 31 is about 61.3º.   b. Graph y = 70. Then use the intersect feature under the CALC menu to determine the interval on

which 8.05 cos + 66.95 > 70.

So, 8.05 cos + 66.95 > 70 on the interval

3.74 < x < 8.26. Because x = 3.74 corresponds to about the middle of April, the temperature does not exceed 70º for the entire month until May. Therefore, the average daily temperature is greater than 70º for the entire month during the months of May, June, and July. Therefore, the average temperature was higher than 70º for three months.   c. Use the maximum feature under the CALC menu to find the maximum value of the function on [0, 12].

The maximum value is 75 when x ≈ 6. Therefore, the maximum average monthly temperature is 75º during the month of June.

Find the x-intercepts of each graph on the interval [0, 2 ).

41. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) cos x = 0 when x =  and x

= .

42. 

 

SOLUTION:  Let y = 0 and solve for x.

  On the interval [0, 2π) sin x = 0 when x = 0 and x = π.

43. 

SOLUTION:  Let y = 0 and solve for x.

 

On the interval [0, 2π) cot x = 1 when x =  and x

= .

44. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) tan x = 0 when x = 0 and x =

π, tan x = –1 when x =  and x = , and tan x

= 1 when x =  and x = .

Find all solutions of each equation on the interval [0, 4 ).

45. 4 tan x = 2 sec2 x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

46. 2 sin2 x + 1 = –3 sin x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, , , and .

47. csc x cot2 x = csc x

SOLUTION:   

On the interval [0, 4π) the solutions are , ,

, , , , , and .

48. sec x + 5 = 2 sec x + 3

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

49. GEOMETRY Consider the circle below.

a. The length s of  is given by s = r(2 ) where 0≤   ≤  . When s = 18 and AB = 14, the radius is r

= . Use a graphing calculator to find the

measure of 2 in radians. b. The area of the shaded region is given by A =

. Use a graphing calculator to find the

radian measure of θ  if the radius is 5 inches and the area is 36 square inches. Round to the nearest hundredth.

SOLUTION:  a. Rewrite the arclength formula using s = 18 and r

= .

On the graphing calculator, find the intersection of Y1 = 18sinθ and Y2 = 14θ.

The value of 2θ = 2(1.1968) or about 2.39 radians.     b. First, substitute into the given area formula and rearrange it.

  Using a graphing calculator, find the intersection of Y1 = 2.88 and Y2 = θ – sin θ.

When the area is 36 square inches and the radius is 5 inches, then the measure of θ is 3.01 radians.

Solve each inequality on the interval [0, 2 ).50. 1 > 2 sin x

SOLUTION:  Graph y = 1 and y = 2 sin x on [0, 2π). Use the intersect feature under the CALC menu to determine on what interval(s) 1 > 2 sin x.

  The graphs intersect at about 0.524 or  and about 

2.618 or . The graph of y = 1 is above the other

graph when x < and when x > . Therefore, 1

> 2 sin x on 0 ≤ x <    or   < x < 2π.

51. 0 < 2 cos x –

SOLUTION:  Graph y = 2 cos x –  on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0 < 2 cos x – .

  The zeros of y = 2 cos x –  are about 0.785 or  and about 5.498 or  . The graph of y = 0 is

below the other graph when 0 ≤ x <  or   < x <

2π. Therefore, 0 < 2 cos x –  on 0 ≤ x <  or   < 

x < 2π.

52.  ≥ 

SOLUTION:  

Graph y = and y =  on [0, 2π). Use

the intersect feature under the CALC menu to

determine on what interval(s) ≥  .

  The graphs intersect at about 1.047 or  and about 

2.094 or . The graph of y = is below the

other graph when   ≤ x ≤  .

Therefore, ≥   on  ≤ x ≤  .

53.  ≤ tan x cot x

SOLUTION:  

Graph y = and y = tan x cot x on [0, 2π).

Use the intersect feature under the CALC menu to

determine on what interval(s) ≤ tan x cot

x.  

  The graphs intersect at about 3.142 or π. The graph of y = tan x cot x is at or below the other graph when 0 ≤ x < 2π.  

Therefore, ≤ tan x cot x on 0 ≤ x < 2π.

54. cos x ≤

SOLUTION:  

Graph y = cos x and y =  on [0, 2π). Use the

intersect feature under the CALC menu to

determine on what interval(s) cos x ≤ .

  The graphs intersect at about 2.618 or  and about

3.665 or . The graph of y = cos x is below the

other graph when  ≤ x ≤  .

 

Therefore, cos x ≤ on  ≤ x ≤  .

55.  sin x − 1 < 0

SOLUTION:  Graph y =  sin x − 1. Use the zero feature underthe CALC menu to determine on what interval(s)

 sin x − 1 < 0.

  The zeros of  sin x − 1 are about 0.785 or  and

about 2.356 or . The graph is below the x-axis

when 0 ≤ x <  or   < x < 2π.

Therefore,  sin x − 1 < 0 on 0 ≤ x <  or   < 

x < 2π.

56. REFRACTION When light travels from one medium to another it bends or refracts, as shown.  

Refraction is described by n2 sinθ1 = n1 sinθ2, wheren1 is the index of refraction of the medium the light

is entering, n2 is the index of refraction of the medium the light is exiting, θ1 is the angle of

incidence, and θ2 is the angle of refraction. a. Find θ2 for each material shown if the angle of incidence is 40º and the index of refraction for air is 1.00.

b. If the angle of incidence is doubled to 80 , will the resulting angles of refraction be twice as large asthose found in part a?

SOLUTION:  a. glass:

  ice:

  plastic:

  water:

  b. Find the angle of refraction for one of the mediums using an angle of incidence of 80 . glass:

θ2 ≈ 25 when θ1 = 40 andθ2 ≈ 40.4 when θ1 = 80 . Therefore, the angle of refraction is not twice as large.

57. ERROR ANALYSIS Vijay and Alicia are solving tan2 x – tan x +  =   tan x. Vijay thinks that

the solutions are x =  + n , x =  + n , x =

 + n , and x =  + n . Alicia thinks that the

solutions are x =  + n and x =  + n . Is

either of them correct? Explain your reasoning.

SOLUTION:  First, solve tan2 x – tan x +  =   tan x.

  On [0, 2π) tan x = 1 when x =  and x =  and 

tan x = when x =  and x = .

  Sample answer: Therefore, Vijay’s solutions are correct; however, they are not stated in the simplest

form. For example, his solutions of x =  + n and

x =  + n could simply be stated as x =  + 

n because when n = 1,  + nπ  is equivalent to

.

CHALLENGE  Solve each equation on the interval [0, 2π].

58. 16 sin5 x + 2 sin x = 12 sin3 x

SOLUTION:  

On [0, 2π) sin x = 0 when x = 0 and x = π, sin x =

when x =  and x = , sin x = when

x =  and x = , sin x =  – when x =  and 

x = , and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are 0, , , , , , , ,

, and .

59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3

SOLUTION:  

On [0, 2 ) sin x = 1 when x = , sin x  = –1 when

x = , sin x =  – when x =  and x = ,

and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are , , , , , and .

60. REASONING Are the solutions of csc x =  andcot2 x + 1 = 2 equivalent? If so, verify your answer algebraically. If not, explain your reasoning.

SOLUTION:  

 

 

The solutions to csc x =  are   and  . Using

a Pythagorean identity, cot2 x + 1 = 2 simplifies to csc2 x = 2 or csc x = ± . Therefore, there are

two additional solutions when csc x = – :  and

. So, the solutions of csc x =  and cot2 x + 1

= 2 are not equivalent.

OPEN ENDED Write a trigonometric equation that has each of the following solutions.

61. 

SOLUTION:  Sample answer: When sin x = 0, x = 0 and x = π.

When sin x = , x =  and x = . So, one

equation that has solutions of 0, π, , and  is 

= 0 or sin2 x –  = 0. This

can be rewritten as 2 sin2 x =  sin x.

62. 

SOLUTION:  

Sample answer: When cos x = , x =  and x =

. When cos x = – , x =  and x = .

So, one equation that has solutions of , , ,

and  is  = 0 or cos2 x

–  = 0. This can be rewritten as 4 cos2 x = 3.

63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying identities.

SOLUTION:  When solving an equation, you use properties of equality to manipulate each side of the equation to isolate a variable. For example:

  When verifying an identity, you transform an expression on one side of the identity into the expression on the other side through a series of algebraic steps.For example:

Verify = .  

Verify each identity.

64.  =

SOLUTION:  

65.  =

SOLUTION:  

66.  +  = 1

SOLUTION:  

+  

= cos2 θ + sin2 θ            Reciprocal Identities= 1                               Pythagorean Identity

Find the value of each expression using the given information.

67. tan ; sin θ = , tan  > 0

SOLUTION:  Use the Pythagorean Identity that involves sin θ.

Since tan  is positive and sin  is positive, cos θ

must be positive. So, .

eSolutions Manual - Powered by Cognero Page 5

5-3 Solving Trigonometric Equations

Page 6: 5-3 Solving Trigonometric Equations the interval , the only solution is when x = 1.34. 39. x log x + 5x cos x = –2 SOLUTION: On the interval , the solutions are when x = 1.84 and

Solve each equation for all values of x.1. 5 sin x + 2 = sin x

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is  + 2nπ,  + 2nπ, .

2. 5 =  sec 2 x + 3

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

3. 2 = 4 cos2 x + 1

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on

the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

4. 4 tan x – 7 = 3 tan x – 6

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

5. 9 + cot2 x = 12

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

6. 2 – 10 sec x = 4 – 9 sec x

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

7. 3 csc x = 2 csc x +

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

8. 11 =  3 csc2 x + 7

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are , , , and . Solutions

on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

9. 6 tan2 x – 2 = 4

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

10. 9 + sin2 x = 10

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the interval

(– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the solutions is

 + 2nπ,  + 2nπ, .

11. 7 cot x –   = 4 cot x

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

12. 7 cos x = 5 cos x +

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ,   + 2nπ, .

Find all solutions of each equation on [0, 2 ).13. sin4 x + 2 sin2 x − 3 = 0

SOLUTION:  

On the interval [0, 2π),  when x =  and 

 when x = . Since  is not a real 

number, the equation  yields no additional solutions.

14. –2 sin x = –sin x cos x

SOLUTION:  

The equation cos x = 2 has no real solutions since the maximum value the cosine function can obtain is 1. On the interval [0, 2π), the equation  has solutions 0 and π.

15. 4 cot x = cot x sin2 x

SOLUTION:  

The equations sin x = 2 and sin x = –2 have no real solutions. On the interval [0, 2π), the equation cot x =

0 has solutions  and  .

16. csc2 x – csc x + 9 = 11

SOLUTION:  

On the interval [0, 2π), the equation csc x = –1 has a

solution of  and the equation csc x = 2 has

solutions of    and  .

17. cos3 x + cos2 x – cos x = 1

SOLUTION:  

On the interval [0, 2π), the equation cos x = 1 has a solution of 0 and the equation cos x = –1 has a solution of  π.

18. 2 sin2 x = sin x + 1

SOLUTION:  

On the interval [0, 2π), the equation sin x = 1 has a

solution of  and the equation sin x =  has  

solutions of    and  .

19. TENNIS A tennis ball leaves a racquet and heads toward a net 40 feet away. The height of the net is the same height as the initial height of the tennis ball. 

a. If the ball is hit at 50 feet per second, neglecting air resistance, use  to find the interval of possible angles of the ball needed to clear the net. b. Find  if the initial velocity remained the same but the distance to the net was 50 feet.

SOLUTION:  a.

The interval is [15.4°, 74.6°].   b.

If the distance to the net is 50 feet, then the angle would be 19.9° or 70.1°.

20. SKIING In the Olympics aerial skiing competition, skiers speed down a slope that launches them into the air, as shown. The maximum height a skier can

reach is given by hpeak =  , where g is

the acceleration due to gravity or 9.8 meters per second squared.

a. If a skier obtains a height of 5 meters above the end of the ramp, what was the skier’s initial speed? b. Use your answer from part a to determine how long it took the skier to reach the maximum height if tpeak = .

SOLUTION:  a.

The skier’s initial speed was 11.67 meters per second.   b.

  The time it took the skier to reach the maximum height was about 1.0 seconds.

Find all solutions of each equation on the interval [0, 2 ).

21. 1 = cot 2 x + csc x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

22. sec x = tan x + 1

SOLUTION:  

   

  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is 0.

23. tan 2 x = 1 – sec x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are 0,

, and .

24. csc x + cot x = 1

SOLUTION:  

 

x =  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is .

25. 2 – 2 cos2 x = sin x + 1

SOLUTION:  

Check

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

26. cos x – 4 = sin x – 4

SOLUTION:   

 

Therefore, on the interval [0, 2π) the only valid

solutions are and .

27. 3 sin x = 3 – 3 cos x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the only valid

solutions are and 0.

28. cot2 x csc2 x – cot2 x = 9

SOLUTION:  

is undefined. The solutions of  are 

  Check

 

 

 

On the interval , there are no real values of x for which , but  for 

29. sec2 x – 1 + tan x –  tan x =

SOLUTION:  

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

30. sec2 x tan2 x + 3 sec2 x – 2 tan2 x = 3

SOLUTION:  

 

The square of any real number must be greater than or equal to zero so sec2 x = –1 has no solutions. Therefore, on the interval [0, 2π) the only solutions are 0 and π.

31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive power PR of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical components, PH and PV.

Using the equations above, determine for what values of  PV and PH are equivalent.

SOLUTION:  

The sine and cosine have the same values in the

interval [0, 2π) at  and  .

Therefore, the components will be equivalent when

.

Find all solutions of each equation on the interval [0, 2 ).

32.  + cos x = 2

SOLUTION:  

On the interval [0, 2π), cos x =  when x =  and 

when x = .

33.  +  = –4

SOLUTION:  

On the interval [0, 2π),cos x = – when x =

 and when x = .

34.  +  = cos x

SOLUTION:  

On the interval [0, 2π) there are no values of x for

which sin x = –2 and the sin x = 1 only when x = .

However, tan  is undefined, so there are no 

solutions for the original equation.

35. cot x cos x + 1 =  + 

SOLUTION:  

On ,  when x =  and when x =

.

GRAPHING CALCULATOR Solve each equation on the interval [0, 2 ) by graphing. Round to the nearest hundredth.

36. 3 cos 2x = ex + 1

SOLUTION:  

On the interval , the only solution is when x = 0.33.

37. sin x + cos x = 3x

SOLUTION:  

On the interval , the only solution is when x = 0.41.

38. x2 = 2 cos x + x

SOLUTION:  

On the interval , the only solution is when x = 1.34.

39. x log x + 5x cos x = –2

SOLUTION:  

On the interval , the solutions are when x = 1.84 and when x = 4.49.

40. METEOROLOGY The average daily temperature in degrees Fahrenheit for a city can be modeled by t

= 8.05 cos + 66.95, where x is a function of

time, x = 1 represents January 15, x = 2 represents February 15, and so on. a. Use a graphing calculator to estimate the temperature on January 31. b. Approximate the number of months that the average daily temperature is greater than 70 throughout the entire month. c. Estimate the highest temperature of the year and the month in which it occurs.

SOLUTION:  a. Use a graphing calculator to graph t = 8.05

cos + 66.95. Because x = 1 represents

January 15, x = 1.5 represents January 31. Use the TRACE feature to determine the temperature whenx = 1.5.

Therefore, the temperature on January 31 is about 61.3º.   b. Graph y = 70. Then use the intersect feature under the CALC menu to determine the interval on

which 8.05 cos + 66.95 > 70.

So, 8.05 cos + 66.95 > 70 on the interval

3.74 < x < 8.26. Because x = 3.74 corresponds to about the middle of April, the temperature does not exceed 70º for the entire month until May. Therefore, the average daily temperature is greater than 70º for the entire month during the months of May, June, and July. Therefore, the average temperature was higher than 70º for three months.   c. Use the maximum feature under the CALC menu to find the maximum value of the function on [0, 12].

The maximum value is 75 when x ≈ 6. Therefore, the maximum average monthly temperature is 75º during the month of June.

Find the x-intercepts of each graph on the interval [0, 2 ).

41. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) cos x = 0 when x =  and x

= .

42. 

 

SOLUTION:  Let y = 0 and solve for x.

  On the interval [0, 2π) sin x = 0 when x = 0 and x = π.

43. 

SOLUTION:  Let y = 0 and solve for x.

 

On the interval [0, 2π) cot x = 1 when x =  and x

= .

44. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) tan x = 0 when x = 0 and x =

π, tan x = –1 when x =  and x = , and tan x

= 1 when x =  and x = .

Find all solutions of each equation on the interval [0, 4 ).

45. 4 tan x = 2 sec2 x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

46. 2 sin2 x + 1 = –3 sin x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, , , and .

47. csc x cot2 x = csc x

SOLUTION:   

On the interval [0, 4π) the solutions are , ,

, , , , , and .

48. sec x + 5 = 2 sec x + 3

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

49. GEOMETRY Consider the circle below.

a. The length s of  is given by s = r(2 ) where 0≤   ≤  . When s = 18 and AB = 14, the radius is r

= . Use a graphing calculator to find the

measure of 2 in radians. b. The area of the shaded region is given by A =

. Use a graphing calculator to find the

radian measure of θ  if the radius is 5 inches and the area is 36 square inches. Round to the nearest hundredth.

SOLUTION:  a. Rewrite the arclength formula using s = 18 and r

= .

On the graphing calculator, find the intersection of Y1 = 18sinθ and Y2 = 14θ.

The value of 2θ = 2(1.1968) or about 2.39 radians.     b. First, substitute into the given area formula and rearrange it.

  Using a graphing calculator, find the intersection of Y1 = 2.88 and Y2 = θ – sin θ.

When the area is 36 square inches and the radius is 5 inches, then the measure of θ is 3.01 radians.

Solve each inequality on the interval [0, 2 ).50. 1 > 2 sin x

SOLUTION:  Graph y = 1 and y = 2 sin x on [0, 2π). Use the intersect feature under the CALC menu to determine on what interval(s) 1 > 2 sin x.

  The graphs intersect at about 0.524 or  and about 

2.618 or . The graph of y = 1 is above the other

graph when x < and when x > . Therefore, 1

> 2 sin x on 0 ≤ x <    or   < x < 2π.

51. 0 < 2 cos x –

SOLUTION:  Graph y = 2 cos x –  on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0 < 2 cos x – .

  The zeros of y = 2 cos x –  are about 0.785 or  and about 5.498 or  . The graph of y = 0 is

below the other graph when 0 ≤ x <  or   < x <

2π. Therefore, 0 < 2 cos x –  on 0 ≤ x <  or   < 

x < 2π.

52.  ≥ 

SOLUTION:  

Graph y = and y =  on [0, 2π). Use

the intersect feature under the CALC menu to

determine on what interval(s) ≥  .

  The graphs intersect at about 1.047 or  and about 

2.094 or . The graph of y = is below the

other graph when   ≤ x ≤  .

Therefore, ≥   on  ≤ x ≤  .

53.  ≤ tan x cot x

SOLUTION:  

Graph y = and y = tan x cot x on [0, 2π).

Use the intersect feature under the CALC menu to

determine on what interval(s) ≤ tan x cot

x.  

  The graphs intersect at about 3.142 or π. The graph of y = tan x cot x is at or below the other graph when 0 ≤ x < 2π.  

Therefore, ≤ tan x cot x on 0 ≤ x < 2π.

54. cos x ≤

SOLUTION:  

Graph y = cos x and y =  on [0, 2π). Use the

intersect feature under the CALC menu to

determine on what interval(s) cos x ≤ .

  The graphs intersect at about 2.618 or  and about

3.665 or . The graph of y = cos x is below the

other graph when  ≤ x ≤  .

 

Therefore, cos x ≤ on  ≤ x ≤  .

55.  sin x − 1 < 0

SOLUTION:  Graph y =  sin x − 1. Use the zero feature underthe CALC menu to determine on what interval(s)

 sin x − 1 < 0.

  The zeros of  sin x − 1 are about 0.785 or  and

about 2.356 or . The graph is below the x-axis

when 0 ≤ x <  or   < x < 2π.

Therefore,  sin x − 1 < 0 on 0 ≤ x <  or   < 

x < 2π.

56. REFRACTION When light travels from one medium to another it bends or refracts, as shown.  

Refraction is described by n2 sinθ1 = n1 sinθ2, wheren1 is the index of refraction of the medium the light

is entering, n2 is the index of refraction of the medium the light is exiting, θ1 is the angle of

incidence, and θ2 is the angle of refraction. a. Find θ2 for each material shown if the angle of incidence is 40º and the index of refraction for air is 1.00.

b. If the angle of incidence is doubled to 80 , will the resulting angles of refraction be twice as large asthose found in part a?

SOLUTION:  a. glass:

  ice:

  plastic:

  water:

  b. Find the angle of refraction for one of the mediums using an angle of incidence of 80 . glass:

θ2 ≈ 25 when θ1 = 40 andθ2 ≈ 40.4 when θ1 = 80 . Therefore, the angle of refraction is not twice as large.

57. ERROR ANALYSIS Vijay and Alicia are solving tan2 x – tan x +  =   tan x. Vijay thinks that

the solutions are x =  + n , x =  + n , x =

 + n , and x =  + n . Alicia thinks that the

solutions are x =  + n and x =  + n . Is

either of them correct? Explain your reasoning.

SOLUTION:  First, solve tan2 x – tan x +  =   tan x.

  On [0, 2π) tan x = 1 when x =  and x =  and 

tan x = when x =  and x = .

  Sample answer: Therefore, Vijay’s solutions are correct; however, they are not stated in the simplest

form. For example, his solutions of x =  + n and

x =  + n could simply be stated as x =  + 

n because when n = 1,  + nπ  is equivalent to

.

CHALLENGE  Solve each equation on the interval [0, 2π].

58. 16 sin5 x + 2 sin x = 12 sin3 x

SOLUTION:  

On [0, 2π) sin x = 0 when x = 0 and x = π, sin x =

when x =  and x = , sin x = when

x =  and x = , sin x =  – when x =  and 

x = , and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are 0, , , , , , , ,

, and .

59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3

SOLUTION:  

On [0, 2 ) sin x = 1 when x = , sin x  = –1 when

x = , sin x =  – when x =  and x = ,

and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are , , , , , and .

60. REASONING Are the solutions of csc x =  andcot2 x + 1 = 2 equivalent? If so, verify your answer algebraically. If not, explain your reasoning.

SOLUTION:  

 

 

The solutions to csc x =  are   and  . Using

a Pythagorean identity, cot2 x + 1 = 2 simplifies to csc2 x = 2 or csc x = ± . Therefore, there are

two additional solutions when csc x = – :  and

. So, the solutions of csc x =  and cot2 x + 1

= 2 are not equivalent.

OPEN ENDED Write a trigonometric equation that has each of the following solutions.

61. 

SOLUTION:  Sample answer: When sin x = 0, x = 0 and x = π.

When sin x = , x =  and x = . So, one

equation that has solutions of 0, π, , and  is 

= 0 or sin2 x –  = 0. This

can be rewritten as 2 sin2 x =  sin x.

62. 

SOLUTION:  

Sample answer: When cos x = , x =  and x =

. When cos x = – , x =  and x = .

So, one equation that has solutions of , , ,

and  is  = 0 or cos2 x

–  = 0. This can be rewritten as 4 cos2 x = 3.

63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying identities.

SOLUTION:  When solving an equation, you use properties of equality to manipulate each side of the equation to isolate a variable. For example:

  When verifying an identity, you transform an expression on one side of the identity into the expression on the other side through a series of algebraic steps.For example:

Verify = .  

Verify each identity.

64.  =

SOLUTION:  

65.  =

SOLUTION:  

66.  +  = 1

SOLUTION:  

+  

= cos2 θ + sin2 θ            Reciprocal Identities= 1                               Pythagorean Identity

Find the value of each expression using the given information.

67. tan ; sin θ = , tan  > 0

SOLUTION:  Use the Pythagorean Identity that involves sin θ.

Since tan  is positive and sin  is positive, cos θ

must be positive. So, .

Solve each equation for all values of x.1. 5 sin x + 2 = sin x

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is  + 2nπ,  + 2nπ, .

2. 5 =  sec 2 x + 3

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

3. 2 = 4 cos2 x + 1

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on

the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

4. 4 tan x – 7 = 3 tan x – 6

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

5. 9 + cot2 x = 12

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

6. 2 – 10 sec x = 4 – 9 sec x

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

7. 3 csc x = 2 csc x +

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

8. 11 =  3 csc2 x + 7

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are , , , and . Solutions

on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

9. 6 tan2 x – 2 = 4

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

10. 9 + sin2 x = 10

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the interval

(– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the solutions is

 + 2nπ,  + 2nπ, .

11. 7 cot x –   = 4 cot x

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

12. 7 cos x = 5 cos x +

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ,   + 2nπ, .

Find all solutions of each equation on [0, 2 ).13. sin4 x + 2 sin2 x − 3 = 0

SOLUTION:  

On the interval [0, 2π),  when x =  and 

 when x = . Since  is not a real 

number, the equation  yields no additional solutions.

14. –2 sin x = –sin x cos x

SOLUTION:  

The equation cos x = 2 has no real solutions since the maximum value the cosine function can obtain is 1. On the interval [0, 2π), the equation  has solutions 0 and π.

15. 4 cot x = cot x sin2 x

SOLUTION:  

The equations sin x = 2 and sin x = –2 have no real solutions. On the interval [0, 2π), the equation cot x =

0 has solutions  and  .

16. csc2 x – csc x + 9 = 11

SOLUTION:  

On the interval [0, 2π), the equation csc x = –1 has a

solution of  and the equation csc x = 2 has

solutions of    and  .

17. cos3 x + cos2 x – cos x = 1

SOLUTION:  

On the interval [0, 2π), the equation cos x = 1 has a solution of 0 and the equation cos x = –1 has a solution of  π.

18. 2 sin2 x = sin x + 1

SOLUTION:  

On the interval [0, 2π), the equation sin x = 1 has a

solution of  and the equation sin x =  has  

solutions of    and  .

19. TENNIS A tennis ball leaves a racquet and heads toward a net 40 feet away. The height of the net is the same height as the initial height of the tennis ball. 

a. If the ball is hit at 50 feet per second, neglecting air resistance, use  to find the interval of possible angles of the ball needed to clear the net. b. Find  if the initial velocity remained the same but the distance to the net was 50 feet.

SOLUTION:  a.

The interval is [15.4°, 74.6°].   b.

If the distance to the net is 50 feet, then the angle would be 19.9° or 70.1°.

20. SKIING In the Olympics aerial skiing competition, skiers speed down a slope that launches them into the air, as shown. The maximum height a skier can

reach is given by hpeak =  , where g is

the acceleration due to gravity or 9.8 meters per second squared.

a. If a skier obtains a height of 5 meters above the end of the ramp, what was the skier’s initial speed? b. Use your answer from part a to determine how long it took the skier to reach the maximum height if tpeak = .

SOLUTION:  a.

The skier’s initial speed was 11.67 meters per second.   b.

  The time it took the skier to reach the maximum height was about 1.0 seconds.

Find all solutions of each equation on the interval [0, 2 ).

21. 1 = cot 2 x + csc x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

22. sec x = tan x + 1

SOLUTION:  

   

  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is 0.

23. tan 2 x = 1 – sec x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are 0,

, and .

24. csc x + cot x = 1

SOLUTION:  

 

x =  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is .

25. 2 – 2 cos2 x = sin x + 1

SOLUTION:  

Check

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

26. cos x – 4 = sin x – 4

SOLUTION:   

 

Therefore, on the interval [0, 2π) the only valid

solutions are and .

27. 3 sin x = 3 – 3 cos x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the only valid

solutions are and 0.

28. cot2 x csc2 x – cot2 x = 9

SOLUTION:  

is undefined. The solutions of  are 

  Check

 

 

 

On the interval , there are no real values of x for which , but  for 

29. sec2 x – 1 + tan x –  tan x =

SOLUTION:  

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

30. sec2 x tan2 x + 3 sec2 x – 2 tan2 x = 3

SOLUTION:  

 

The square of any real number must be greater than or equal to zero so sec2 x = –1 has no solutions. Therefore, on the interval [0, 2π) the only solutions are 0 and π.

31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive power PR of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical components, PH and PV.

Using the equations above, determine for what values of  PV and PH are equivalent.

SOLUTION:  

The sine and cosine have the same values in the

interval [0, 2π) at  and  .

Therefore, the components will be equivalent when

.

Find all solutions of each equation on the interval [0, 2 ).

32.  + cos x = 2

SOLUTION:  

On the interval [0, 2π), cos x =  when x =  and 

when x = .

33.  +  = –4

SOLUTION:  

On the interval [0, 2π),cos x = – when x =

 and when x = .

34.  +  = cos x

SOLUTION:  

On the interval [0, 2π) there are no values of x for

which sin x = –2 and the sin x = 1 only when x = .

However, tan  is undefined, so there are no 

solutions for the original equation.

35. cot x cos x + 1 =  + 

SOLUTION:  

On ,  when x =  and when x =

.

GRAPHING CALCULATOR Solve each equation on the interval [0, 2 ) by graphing. Round to the nearest hundredth.

36. 3 cos 2x = ex + 1

SOLUTION:  

On the interval , the only solution is when x = 0.33.

37. sin x + cos x = 3x

SOLUTION:  

On the interval , the only solution is when x = 0.41.

38. x2 = 2 cos x + x

SOLUTION:  

On the interval , the only solution is when x = 1.34.

39. x log x + 5x cos x = –2

SOLUTION:  

On the interval , the solutions are when x = 1.84 and when x = 4.49.

40. METEOROLOGY The average daily temperature in degrees Fahrenheit for a city can be modeled by t

= 8.05 cos + 66.95, where x is a function of

time, x = 1 represents January 15, x = 2 represents February 15, and so on. a. Use a graphing calculator to estimate the temperature on January 31. b. Approximate the number of months that the average daily temperature is greater than 70 throughout the entire month. c. Estimate the highest temperature of the year and the month in which it occurs.

SOLUTION:  a. Use a graphing calculator to graph t = 8.05

cos + 66.95. Because x = 1 represents

January 15, x = 1.5 represents January 31. Use the TRACE feature to determine the temperature whenx = 1.5.

Therefore, the temperature on January 31 is about 61.3º.   b. Graph y = 70. Then use the intersect feature under the CALC menu to determine the interval on

which 8.05 cos + 66.95 > 70.

So, 8.05 cos + 66.95 > 70 on the interval

3.74 < x < 8.26. Because x = 3.74 corresponds to about the middle of April, the temperature does not exceed 70º for the entire month until May. Therefore, the average daily temperature is greater than 70º for the entire month during the months of May, June, and July. Therefore, the average temperature was higher than 70º for three months.   c. Use the maximum feature under the CALC menu to find the maximum value of the function on [0, 12].

The maximum value is 75 when x ≈ 6. Therefore, the maximum average monthly temperature is 75º during the month of June.

Find the x-intercepts of each graph on the interval [0, 2 ).

41. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) cos x = 0 when x =  and x

= .

42. 

 

SOLUTION:  Let y = 0 and solve for x.

  On the interval [0, 2π) sin x = 0 when x = 0 and x = π.

43. 

SOLUTION:  Let y = 0 and solve for x.

 

On the interval [0, 2π) cot x = 1 when x =  and x

= .

44. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) tan x = 0 when x = 0 and x =

π, tan x = –1 when x =  and x = , and tan x

= 1 when x =  and x = .

Find all solutions of each equation on the interval [0, 4 ).

45. 4 tan x = 2 sec2 x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

46. 2 sin2 x + 1 = –3 sin x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, , , and .

47. csc x cot2 x = csc x

SOLUTION:   

On the interval [0, 4π) the solutions are , ,

, , , , , and .

48. sec x + 5 = 2 sec x + 3

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

49. GEOMETRY Consider the circle below.

a. The length s of  is given by s = r(2 ) where 0≤   ≤  . When s = 18 and AB = 14, the radius is r

= . Use a graphing calculator to find the

measure of 2 in radians. b. The area of the shaded region is given by A =

. Use a graphing calculator to find the

radian measure of θ  if the radius is 5 inches and the area is 36 square inches. Round to the nearest hundredth.

SOLUTION:  a. Rewrite the arclength formula using s = 18 and r

= .

On the graphing calculator, find the intersection of Y1 = 18sinθ and Y2 = 14θ.

The value of 2θ = 2(1.1968) or about 2.39 radians.     b. First, substitute into the given area formula and rearrange it.

  Using a graphing calculator, find the intersection of Y1 = 2.88 and Y2 = θ – sin θ.

When the area is 36 square inches and the radius is 5 inches, then the measure of θ is 3.01 radians.

Solve each inequality on the interval [0, 2 ).50. 1 > 2 sin x

SOLUTION:  Graph y = 1 and y = 2 sin x on [0, 2π). Use the intersect feature under the CALC menu to determine on what interval(s) 1 > 2 sin x.

  The graphs intersect at about 0.524 or  and about 

2.618 or . The graph of y = 1 is above the other

graph when x < and when x > . Therefore, 1

> 2 sin x on 0 ≤ x <    or   < x < 2π.

51. 0 < 2 cos x –

SOLUTION:  Graph y = 2 cos x –  on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0 < 2 cos x – .

  The zeros of y = 2 cos x –  are about 0.785 or  and about 5.498 or  . The graph of y = 0 is

below the other graph when 0 ≤ x <  or   < x <

2π. Therefore, 0 < 2 cos x –  on 0 ≤ x <  or   < 

x < 2π.

52.  ≥ 

SOLUTION:  

Graph y = and y =  on [0, 2π). Use

the intersect feature under the CALC menu to

determine on what interval(s) ≥  .

  The graphs intersect at about 1.047 or  and about 

2.094 or . The graph of y = is below the

other graph when   ≤ x ≤  .

Therefore, ≥   on  ≤ x ≤  .

53.  ≤ tan x cot x

SOLUTION:  

Graph y = and y = tan x cot x on [0, 2π).

Use the intersect feature under the CALC menu to

determine on what interval(s) ≤ tan x cot

x.  

  The graphs intersect at about 3.142 or π. The graph of y = tan x cot x is at or below the other graph when 0 ≤ x < 2π.  

Therefore, ≤ tan x cot x on 0 ≤ x < 2π.

54. cos x ≤

SOLUTION:  

Graph y = cos x and y =  on [0, 2π). Use the

intersect feature under the CALC menu to

determine on what interval(s) cos x ≤ .

  The graphs intersect at about 2.618 or  and about

3.665 or . The graph of y = cos x is below the

other graph when  ≤ x ≤  .

 

Therefore, cos x ≤ on  ≤ x ≤  .

55.  sin x − 1 < 0

SOLUTION:  Graph y =  sin x − 1. Use the zero feature underthe CALC menu to determine on what interval(s)

 sin x − 1 < 0.

  The zeros of  sin x − 1 are about 0.785 or  and

about 2.356 or . The graph is below the x-axis

when 0 ≤ x <  or   < x < 2π.

Therefore,  sin x − 1 < 0 on 0 ≤ x <  or   < 

x < 2π.

56. REFRACTION When light travels from one medium to another it bends or refracts, as shown.  

Refraction is described by n2 sinθ1 = n1 sinθ2, wheren1 is the index of refraction of the medium the light

is entering, n2 is the index of refraction of the medium the light is exiting, θ1 is the angle of

incidence, and θ2 is the angle of refraction. a. Find θ2 for each material shown if the angle of incidence is 40º and the index of refraction for air is 1.00.

b. If the angle of incidence is doubled to 80 , will the resulting angles of refraction be twice as large asthose found in part a?

SOLUTION:  a. glass:

  ice:

  plastic:

  water:

  b. Find the angle of refraction for one of the mediums using an angle of incidence of 80 . glass:

θ2 ≈ 25 when θ1 = 40 andθ2 ≈ 40.4 when θ1 = 80 . Therefore, the angle of refraction is not twice as large.

57. ERROR ANALYSIS Vijay and Alicia are solving tan2 x – tan x +  =   tan x. Vijay thinks that

the solutions are x =  + n , x =  + n , x =

 + n , and x =  + n . Alicia thinks that the

solutions are x =  + n and x =  + n . Is

either of them correct? Explain your reasoning.

SOLUTION:  First, solve tan2 x – tan x +  =   tan x.

  On [0, 2π) tan x = 1 when x =  and x =  and 

tan x = when x =  and x = .

  Sample answer: Therefore, Vijay’s solutions are correct; however, they are not stated in the simplest

form. For example, his solutions of x =  + n and

x =  + n could simply be stated as x =  + 

n because when n = 1,  + nπ  is equivalent to

.

CHALLENGE  Solve each equation on the interval [0, 2π].

58. 16 sin5 x + 2 sin x = 12 sin3 x

SOLUTION:  

On [0, 2π) sin x = 0 when x = 0 and x = π, sin x =

when x =  and x = , sin x = when

x =  and x = , sin x =  – when x =  and 

x = , and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are 0, , , , , , , ,

, and .

59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3

SOLUTION:  

On [0, 2 ) sin x = 1 when x = , sin x  = –1 when

x = , sin x =  – when x =  and x = ,

and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are , , , , , and .

60. REASONING Are the solutions of csc x =  andcot2 x + 1 = 2 equivalent? If so, verify your answer algebraically. If not, explain your reasoning.

SOLUTION:  

 

 

The solutions to csc x =  are   and  . Using

a Pythagorean identity, cot2 x + 1 = 2 simplifies to csc2 x = 2 or csc x = ± . Therefore, there are

two additional solutions when csc x = – :  and

. So, the solutions of csc x =  and cot2 x + 1

= 2 are not equivalent.

OPEN ENDED Write a trigonometric equation that has each of the following solutions.

61. 

SOLUTION:  Sample answer: When sin x = 0, x = 0 and x = π.

When sin x = , x =  and x = . So, one

equation that has solutions of 0, π, , and  is 

= 0 or sin2 x –  = 0. This

can be rewritten as 2 sin2 x =  sin x.

62. 

SOLUTION:  

Sample answer: When cos x = , x =  and x =

. When cos x = – , x =  and x = .

So, one equation that has solutions of , , ,

and  is  = 0 or cos2 x

–  = 0. This can be rewritten as 4 cos2 x = 3.

63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying identities.

SOLUTION:  When solving an equation, you use properties of equality to manipulate each side of the equation to isolate a variable. For example:

  When verifying an identity, you transform an expression on one side of the identity into the expression on the other side through a series of algebraic steps.For example:

Verify = .  

Verify each identity.

64.  =

SOLUTION:  

65.  =

SOLUTION:  

66.  +  = 1

SOLUTION:  

+  

= cos2 θ + sin2 θ            Reciprocal Identities= 1                               Pythagorean Identity

Find the value of each expression using the given information.

67. tan ; sin θ = , tan  > 0

SOLUTION:  Use the Pythagorean Identity that involves sin θ.

Since tan  is positive and sin  is positive, cos θ

must be positive. So, .

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5-3 Solving Trigonometric Equations

Page 7: 5-3 Solving Trigonometric Equations the interval , the only solution is when x = 1.34. 39. x log x + 5x cos x = –2 SOLUTION: On the interval , the solutions are when x = 1.84 and

Solve each equation for all values of x.1. 5 sin x + 2 = sin x

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is  + 2nπ,  + 2nπ, .

2. 5 =  sec 2 x + 3

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

3. 2 = 4 cos2 x + 1

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on

the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

4. 4 tan x – 7 = 3 tan x – 6

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

5. 9 + cot2 x = 12

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

6. 2 – 10 sec x = 4 – 9 sec x

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

7. 3 csc x = 2 csc x +

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

8. 11 =  3 csc2 x + 7

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are , , , and . Solutions

on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

9. 6 tan2 x – 2 = 4

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

10. 9 + sin2 x = 10

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the interval

(– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the solutions is

 + 2nπ,  + 2nπ, .

11. 7 cot x –   = 4 cot x

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

12. 7 cos x = 5 cos x +

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ,   + 2nπ, .

Find all solutions of each equation on [0, 2 ).13. sin4 x + 2 sin2 x − 3 = 0

SOLUTION:  

On the interval [0, 2π),  when x =  and 

 when x = . Since  is not a real 

number, the equation  yields no additional solutions.

14. –2 sin x = –sin x cos x

SOLUTION:  

The equation cos x = 2 has no real solutions since the maximum value the cosine function can obtain is 1. On the interval [0, 2π), the equation  has solutions 0 and π.

15. 4 cot x = cot x sin2 x

SOLUTION:  

The equations sin x = 2 and sin x = –2 have no real solutions. On the interval [0, 2π), the equation cot x =

0 has solutions  and  .

16. csc2 x – csc x + 9 = 11

SOLUTION:  

On the interval [0, 2π), the equation csc x = –1 has a

solution of  and the equation csc x = 2 has

solutions of    and  .

17. cos3 x + cos2 x – cos x = 1

SOLUTION:  

On the interval [0, 2π), the equation cos x = 1 has a solution of 0 and the equation cos x = –1 has a solution of  π.

18. 2 sin2 x = sin x + 1

SOLUTION:  

On the interval [0, 2π), the equation sin x = 1 has a

solution of  and the equation sin x =  has  

solutions of    and  .

19. TENNIS A tennis ball leaves a racquet and heads toward a net 40 feet away. The height of the net is the same height as the initial height of the tennis ball. 

a. If the ball is hit at 50 feet per second, neglecting air resistance, use  to find the interval of possible angles of the ball needed to clear the net. b. Find  if the initial velocity remained the same but the distance to the net was 50 feet.

SOLUTION:  a.

The interval is [15.4°, 74.6°].   b.

If the distance to the net is 50 feet, then the angle would be 19.9° or 70.1°.

20. SKIING In the Olympics aerial skiing competition, skiers speed down a slope that launches them into the air, as shown. The maximum height a skier can

reach is given by hpeak =  , where g is

the acceleration due to gravity or 9.8 meters per second squared.

a. If a skier obtains a height of 5 meters above the end of the ramp, what was the skier’s initial speed? b. Use your answer from part a to determine how long it took the skier to reach the maximum height if tpeak = .

SOLUTION:  a.

The skier’s initial speed was 11.67 meters per second.   b.

  The time it took the skier to reach the maximum height was about 1.0 seconds.

Find all solutions of each equation on the interval [0, 2 ).

21. 1 = cot 2 x + csc x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

22. sec x = tan x + 1

SOLUTION:  

   

  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is 0.

23. tan 2 x = 1 – sec x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are 0,

, and .

24. csc x + cot x = 1

SOLUTION:  

 

x =  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is .

25. 2 – 2 cos2 x = sin x + 1

SOLUTION:  

Check

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

26. cos x – 4 = sin x – 4

SOLUTION:   

 

Therefore, on the interval [0, 2π) the only valid

solutions are and .

27. 3 sin x = 3 – 3 cos x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the only valid

solutions are and 0.

28. cot2 x csc2 x – cot2 x = 9

SOLUTION:  

is undefined. The solutions of  are 

  Check

 

 

 

On the interval , there are no real values of x for which , but  for 

29. sec2 x – 1 + tan x –  tan x =

SOLUTION:  

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

30. sec2 x tan2 x + 3 sec2 x – 2 tan2 x = 3

SOLUTION:  

 

The square of any real number must be greater than or equal to zero so sec2 x = –1 has no solutions. Therefore, on the interval [0, 2π) the only solutions are 0 and π.

31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive power PR of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical components, PH and PV.

Using the equations above, determine for what values of  PV and PH are equivalent.

SOLUTION:  

The sine and cosine have the same values in the

interval [0, 2π) at  and  .

Therefore, the components will be equivalent when

.

Find all solutions of each equation on the interval [0, 2 ).

32.  + cos x = 2

SOLUTION:  

On the interval [0, 2π), cos x =  when x =  and 

when x = .

33.  +  = –4

SOLUTION:  

On the interval [0, 2π),cos x = – when x =

 and when x = .

34.  +  = cos x

SOLUTION:  

On the interval [0, 2π) there are no values of x for

which sin x = –2 and the sin x = 1 only when x = .

However, tan  is undefined, so there are no 

solutions for the original equation.

35. cot x cos x + 1 =  + 

SOLUTION:  

On ,  when x =  and when x =

.

GRAPHING CALCULATOR Solve each equation on the interval [0, 2 ) by graphing. Round to the nearest hundredth.

36. 3 cos 2x = ex + 1

SOLUTION:  

On the interval , the only solution is when x = 0.33.

37. sin x + cos x = 3x

SOLUTION:  

On the interval , the only solution is when x = 0.41.

38. x2 = 2 cos x + x

SOLUTION:  

On the interval , the only solution is when x = 1.34.

39. x log x + 5x cos x = –2

SOLUTION:  

On the interval , the solutions are when x = 1.84 and when x = 4.49.

40. METEOROLOGY The average daily temperature in degrees Fahrenheit for a city can be modeled by t

= 8.05 cos + 66.95, where x is a function of

time, x = 1 represents January 15, x = 2 represents February 15, and so on. a. Use a graphing calculator to estimate the temperature on January 31. b. Approximate the number of months that the average daily temperature is greater than 70 throughout the entire month. c. Estimate the highest temperature of the year and the month in which it occurs.

SOLUTION:  a. Use a graphing calculator to graph t = 8.05

cos + 66.95. Because x = 1 represents

January 15, x = 1.5 represents January 31. Use the TRACE feature to determine the temperature whenx = 1.5.

Therefore, the temperature on January 31 is about 61.3º.   b. Graph y = 70. Then use the intersect feature under the CALC menu to determine the interval on

which 8.05 cos + 66.95 > 70.

So, 8.05 cos + 66.95 > 70 on the interval

3.74 < x < 8.26. Because x = 3.74 corresponds to about the middle of April, the temperature does not exceed 70º for the entire month until May. Therefore, the average daily temperature is greater than 70º for the entire month during the months of May, June, and July. Therefore, the average temperature was higher than 70º for three months.   c. Use the maximum feature under the CALC menu to find the maximum value of the function on [0, 12].

The maximum value is 75 when x ≈ 6. Therefore, the maximum average monthly temperature is 75º during the month of June.

Find the x-intercepts of each graph on the interval [0, 2 ).

41. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) cos x = 0 when x =  and x

= .

42. 

 

SOLUTION:  Let y = 0 and solve for x.

  On the interval [0, 2π) sin x = 0 when x = 0 and x = π.

43. 

SOLUTION:  Let y = 0 and solve for x.

 

On the interval [0, 2π) cot x = 1 when x =  and x

= .

44. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) tan x = 0 when x = 0 and x =

π, tan x = –1 when x =  and x = , and tan x

= 1 when x =  and x = .

Find all solutions of each equation on the interval [0, 4 ).

45. 4 tan x = 2 sec2 x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

46. 2 sin2 x + 1 = –3 sin x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, , , and .

47. csc x cot2 x = csc x

SOLUTION:   

On the interval [0, 4π) the solutions are , ,

, , , , , and .

48. sec x + 5 = 2 sec x + 3

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

49. GEOMETRY Consider the circle below.

a. The length s of  is given by s = r(2 ) where 0≤   ≤  . When s = 18 and AB = 14, the radius is r

= . Use a graphing calculator to find the

measure of 2 in radians. b. The area of the shaded region is given by A =

. Use a graphing calculator to find the

radian measure of θ  if the radius is 5 inches and the area is 36 square inches. Round to the nearest hundredth.

SOLUTION:  a. Rewrite the arclength formula using s = 18 and r

= .

On the graphing calculator, find the intersection of Y1 = 18sinθ and Y2 = 14θ.

The value of 2θ = 2(1.1968) or about 2.39 radians.     b. First, substitute into the given area formula and rearrange it.

  Using a graphing calculator, find the intersection of Y1 = 2.88 and Y2 = θ – sin θ.

When the area is 36 square inches and the radius is 5 inches, then the measure of θ is 3.01 radians.

Solve each inequality on the interval [0, 2 ).50. 1 > 2 sin x

SOLUTION:  Graph y = 1 and y = 2 sin x on [0, 2π). Use the intersect feature under the CALC menu to determine on what interval(s) 1 > 2 sin x.

  The graphs intersect at about 0.524 or  and about 

2.618 or . The graph of y = 1 is above the other

graph when x < and when x > . Therefore, 1

> 2 sin x on 0 ≤ x <    or   < x < 2π.

51. 0 < 2 cos x –

SOLUTION:  Graph y = 2 cos x –  on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0 < 2 cos x – .

  The zeros of y = 2 cos x –  are about 0.785 or  and about 5.498 or  . The graph of y = 0 is

below the other graph when 0 ≤ x <  or   < x <

2π. Therefore, 0 < 2 cos x –  on 0 ≤ x <  or   < 

x < 2π.

52.  ≥ 

SOLUTION:  

Graph y = and y =  on [0, 2π). Use

the intersect feature under the CALC menu to

determine on what interval(s) ≥  .

  The graphs intersect at about 1.047 or  and about 

2.094 or . The graph of y = is below the

other graph when   ≤ x ≤  .

Therefore, ≥   on  ≤ x ≤  .

53.  ≤ tan x cot x

SOLUTION:  

Graph y = and y = tan x cot x on [0, 2π).

Use the intersect feature under the CALC menu to

determine on what interval(s) ≤ tan x cot

x.  

  The graphs intersect at about 3.142 or π. The graph of y = tan x cot x is at or below the other graph when 0 ≤ x < 2π.  

Therefore, ≤ tan x cot x on 0 ≤ x < 2π.

54. cos x ≤

SOLUTION:  

Graph y = cos x and y =  on [0, 2π). Use the

intersect feature under the CALC menu to

determine on what interval(s) cos x ≤ .

  The graphs intersect at about 2.618 or  and about

3.665 or . The graph of y = cos x is below the

other graph when  ≤ x ≤  .

 

Therefore, cos x ≤ on  ≤ x ≤  .

55.  sin x − 1 < 0

SOLUTION:  Graph y =  sin x − 1. Use the zero feature underthe CALC menu to determine on what interval(s)

 sin x − 1 < 0.

  The zeros of  sin x − 1 are about 0.785 or  and

about 2.356 or . The graph is below the x-axis

when 0 ≤ x <  or   < x < 2π.

Therefore,  sin x − 1 < 0 on 0 ≤ x <  or   < 

x < 2π.

56. REFRACTION When light travels from one medium to another it bends or refracts, as shown.  

Refraction is described by n2 sinθ1 = n1 sinθ2, wheren1 is the index of refraction of the medium the light

is entering, n2 is the index of refraction of the medium the light is exiting, θ1 is the angle of

incidence, and θ2 is the angle of refraction. a. Find θ2 for each material shown if the angle of incidence is 40º and the index of refraction for air is 1.00.

b. If the angle of incidence is doubled to 80 , will the resulting angles of refraction be twice as large asthose found in part a?

SOLUTION:  a. glass:

  ice:

  plastic:

  water:

  b. Find the angle of refraction for one of the mediums using an angle of incidence of 80 . glass:

θ2 ≈ 25 when θ1 = 40 andθ2 ≈ 40.4 when θ1 = 80 . Therefore, the angle of refraction is not twice as large.

57. ERROR ANALYSIS Vijay and Alicia are solving tan2 x – tan x +  =   tan x. Vijay thinks that

the solutions are x =  + n , x =  + n , x =

 + n , and x =  + n . Alicia thinks that the

solutions are x =  + n and x =  + n . Is

either of them correct? Explain your reasoning.

SOLUTION:  First, solve tan2 x – tan x +  =   tan x.

  On [0, 2π) tan x = 1 when x =  and x =  and 

tan x = when x =  and x = .

  Sample answer: Therefore, Vijay’s solutions are correct; however, they are not stated in the simplest

form. For example, his solutions of x =  + n and

x =  + n could simply be stated as x =  + 

n because when n = 1,  + nπ  is equivalent to

.

CHALLENGE  Solve each equation on the interval [0, 2π].

58. 16 sin5 x + 2 sin x = 12 sin3 x

SOLUTION:  

On [0, 2π) sin x = 0 when x = 0 and x = π, sin x =

when x =  and x = , sin x = when

x =  and x = , sin x =  – when x =  and 

x = , and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are 0, , , , , , , ,

, and .

59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3

SOLUTION:  

On [0, 2 ) sin x = 1 when x = , sin x  = –1 when

x = , sin x =  – when x =  and x = ,

and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are , , , , , and .

60. REASONING Are the solutions of csc x =  andcot2 x + 1 = 2 equivalent? If so, verify your answer algebraically. If not, explain your reasoning.

SOLUTION:  

 

 

The solutions to csc x =  are   and  . Using

a Pythagorean identity, cot2 x + 1 = 2 simplifies to csc2 x = 2 or csc x = ± . Therefore, there are

two additional solutions when csc x = – :  and

. So, the solutions of csc x =  and cot2 x + 1

= 2 are not equivalent.

OPEN ENDED Write a trigonometric equation that has each of the following solutions.

61. 

SOLUTION:  Sample answer: When sin x = 0, x = 0 and x = π.

When sin x = , x =  and x = . So, one

equation that has solutions of 0, π, , and  is 

= 0 or sin2 x –  = 0. This

can be rewritten as 2 sin2 x =  sin x.

62. 

SOLUTION:  

Sample answer: When cos x = , x =  and x =

. When cos x = – , x =  and x = .

So, one equation that has solutions of , , ,

and  is  = 0 or cos2 x

–  = 0. This can be rewritten as 4 cos2 x = 3.

63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying identities.

SOLUTION:  When solving an equation, you use properties of equality to manipulate each side of the equation to isolate a variable. For example:

  When verifying an identity, you transform an expression on one side of the identity into the expression on the other side through a series of algebraic steps.For example:

Verify = .  

Verify each identity.

64.  =

SOLUTION:  

65.  =

SOLUTION:  

66.  +  = 1

SOLUTION:  

+  

= cos2 θ + sin2 θ            Reciprocal Identities= 1                               Pythagorean Identity

Find the value of each expression using the given information.

67. tan ; sin θ = , tan  > 0

SOLUTION:  Use the Pythagorean Identity that involves sin θ.

Since tan  is positive and sin  is positive, cos θ

must be positive. So, .

Solve each equation for all values of x.1. 5 sin x + 2 = sin x

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is  + 2nπ,  + 2nπ, .

2. 5 =  sec 2 x + 3

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

3. 2 = 4 cos2 x + 1

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on

the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

4. 4 tan x – 7 = 3 tan x – 6

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

5. 9 + cot2 x = 12

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

6. 2 – 10 sec x = 4 – 9 sec x

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

7. 3 csc x = 2 csc x +

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

8. 11 =  3 csc2 x + 7

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are , , , and . Solutions

on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

9. 6 tan2 x – 2 = 4

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

10. 9 + sin2 x = 10

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the interval

(– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the solutions is

 + 2nπ,  + 2nπ, .

11. 7 cot x –   = 4 cot x

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

12. 7 cos x = 5 cos x +

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ,   + 2nπ, .

Find all solutions of each equation on [0, 2 ).13. sin4 x + 2 sin2 x − 3 = 0

SOLUTION:  

On the interval [0, 2π),  when x =  and 

 when x = . Since  is not a real 

number, the equation  yields no additional solutions.

14. –2 sin x = –sin x cos x

SOLUTION:  

The equation cos x = 2 has no real solutions since the maximum value the cosine function can obtain is 1. On the interval [0, 2π), the equation  has solutions 0 and π.

15. 4 cot x = cot x sin2 x

SOLUTION:  

The equations sin x = 2 and sin x = –2 have no real solutions. On the interval [0, 2π), the equation cot x =

0 has solutions  and  .

16. csc2 x – csc x + 9 = 11

SOLUTION:  

On the interval [0, 2π), the equation csc x = –1 has a

solution of  and the equation csc x = 2 has

solutions of    and  .

17. cos3 x + cos2 x – cos x = 1

SOLUTION:  

On the interval [0, 2π), the equation cos x = 1 has a solution of 0 and the equation cos x = –1 has a solution of  π.

18. 2 sin2 x = sin x + 1

SOLUTION:  

On the interval [0, 2π), the equation sin x = 1 has a

solution of  and the equation sin x =  has  

solutions of    and  .

19. TENNIS A tennis ball leaves a racquet and heads toward a net 40 feet away. The height of the net is the same height as the initial height of the tennis ball. 

a. If the ball is hit at 50 feet per second, neglecting air resistance, use  to find the interval of possible angles of the ball needed to clear the net. b. Find  if the initial velocity remained the same but the distance to the net was 50 feet.

SOLUTION:  a.

The interval is [15.4°, 74.6°].   b.

If the distance to the net is 50 feet, then the angle would be 19.9° or 70.1°.

20. SKIING In the Olympics aerial skiing competition, skiers speed down a slope that launches them into the air, as shown. The maximum height a skier can

reach is given by hpeak =  , where g is

the acceleration due to gravity or 9.8 meters per second squared.

a. If a skier obtains a height of 5 meters above the end of the ramp, what was the skier’s initial speed? b. Use your answer from part a to determine how long it took the skier to reach the maximum height if tpeak = .

SOLUTION:  a.

The skier’s initial speed was 11.67 meters per second.   b.

  The time it took the skier to reach the maximum height was about 1.0 seconds.

Find all solutions of each equation on the interval [0, 2 ).

21. 1 = cot 2 x + csc x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

22. sec x = tan x + 1

SOLUTION:  

   

  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is 0.

23. tan 2 x = 1 – sec x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are 0,

, and .

24. csc x + cot x = 1

SOLUTION:  

 

x =  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is .

25. 2 – 2 cos2 x = sin x + 1

SOLUTION:  

Check

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

26. cos x – 4 = sin x – 4

SOLUTION:   

 

Therefore, on the interval [0, 2π) the only valid

solutions are and .

27. 3 sin x = 3 – 3 cos x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the only valid

solutions are and 0.

28. cot2 x csc2 x – cot2 x = 9

SOLUTION:  

is undefined. The solutions of  are 

  Check

 

 

 

On the interval , there are no real values of x for which , but  for 

29. sec2 x – 1 + tan x –  tan x =

SOLUTION:  

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

30. sec2 x tan2 x + 3 sec2 x – 2 tan2 x = 3

SOLUTION:  

 

The square of any real number must be greater than or equal to zero so sec2 x = –1 has no solutions. Therefore, on the interval [0, 2π) the only solutions are 0 and π.

31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive power PR of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical components, PH and PV.

Using the equations above, determine for what values of  PV and PH are equivalent.

SOLUTION:  

The sine and cosine have the same values in the

interval [0, 2π) at  and  .

Therefore, the components will be equivalent when

.

Find all solutions of each equation on the interval [0, 2 ).

32.  + cos x = 2

SOLUTION:  

On the interval [0, 2π), cos x =  when x =  and 

when x = .

33.  +  = –4

SOLUTION:  

On the interval [0, 2π),cos x = – when x =

 and when x = .

34.  +  = cos x

SOLUTION:  

On the interval [0, 2π) there are no values of x for

which sin x = –2 and the sin x = 1 only when x = .

However, tan  is undefined, so there are no 

solutions for the original equation.

35. cot x cos x + 1 =  + 

SOLUTION:  

On ,  when x =  and when x =

.

GRAPHING CALCULATOR Solve each equation on the interval [0, 2 ) by graphing. Round to the nearest hundredth.

36. 3 cos 2x = ex + 1

SOLUTION:  

On the interval , the only solution is when x = 0.33.

37. sin x + cos x = 3x

SOLUTION:  

On the interval , the only solution is when x = 0.41.

38. x2 = 2 cos x + x

SOLUTION:  

On the interval , the only solution is when x = 1.34.

39. x log x + 5x cos x = –2

SOLUTION:  

On the interval , the solutions are when x = 1.84 and when x = 4.49.

40. METEOROLOGY The average daily temperature in degrees Fahrenheit for a city can be modeled by t

= 8.05 cos + 66.95, where x is a function of

time, x = 1 represents January 15, x = 2 represents February 15, and so on. a. Use a graphing calculator to estimate the temperature on January 31. b. Approximate the number of months that the average daily temperature is greater than 70 throughout the entire month. c. Estimate the highest temperature of the year and the month in which it occurs.

SOLUTION:  a. Use a graphing calculator to graph t = 8.05

cos + 66.95. Because x = 1 represents

January 15, x = 1.5 represents January 31. Use the TRACE feature to determine the temperature whenx = 1.5.

Therefore, the temperature on January 31 is about 61.3º.   b. Graph y = 70. Then use the intersect feature under the CALC menu to determine the interval on

which 8.05 cos + 66.95 > 70.

So, 8.05 cos + 66.95 > 70 on the interval

3.74 < x < 8.26. Because x = 3.74 corresponds to about the middle of April, the temperature does not exceed 70º for the entire month until May. Therefore, the average daily temperature is greater than 70º for the entire month during the months of May, June, and July. Therefore, the average temperature was higher than 70º for three months.   c. Use the maximum feature under the CALC menu to find the maximum value of the function on [0, 12].

The maximum value is 75 when x ≈ 6. Therefore, the maximum average monthly temperature is 75º during the month of June.

Find the x-intercepts of each graph on the interval [0, 2 ).

41. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) cos x = 0 when x =  and x

= .

42. 

 

SOLUTION:  Let y = 0 and solve for x.

  On the interval [0, 2π) sin x = 0 when x = 0 and x = π.

43. 

SOLUTION:  Let y = 0 and solve for x.

 

On the interval [0, 2π) cot x = 1 when x =  and x

= .

44. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) tan x = 0 when x = 0 and x =

π, tan x = –1 when x =  and x = , and tan x

= 1 when x =  and x = .

Find all solutions of each equation on the interval [0, 4 ).

45. 4 tan x = 2 sec2 x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

46. 2 sin2 x + 1 = –3 sin x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, , , and .

47. csc x cot2 x = csc x

SOLUTION:   

On the interval [0, 4π) the solutions are , ,

, , , , , and .

48. sec x + 5 = 2 sec x + 3

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

49. GEOMETRY Consider the circle below.

a. The length s of  is given by s = r(2 ) where 0≤   ≤  . When s = 18 and AB = 14, the radius is r

= . Use a graphing calculator to find the

measure of 2 in radians. b. The area of the shaded region is given by A =

. Use a graphing calculator to find the

radian measure of θ  if the radius is 5 inches and the area is 36 square inches. Round to the nearest hundredth.

SOLUTION:  a. Rewrite the arclength formula using s = 18 and r

= .

On the graphing calculator, find the intersection of Y1 = 18sinθ and Y2 = 14θ.

The value of 2θ = 2(1.1968) or about 2.39 radians.     b. First, substitute into the given area formula and rearrange it.

  Using a graphing calculator, find the intersection of Y1 = 2.88 and Y2 = θ – sin θ.

When the area is 36 square inches and the radius is 5 inches, then the measure of θ is 3.01 radians.

Solve each inequality on the interval [0, 2 ).50. 1 > 2 sin x

SOLUTION:  Graph y = 1 and y = 2 sin x on [0, 2π). Use the intersect feature under the CALC menu to determine on what interval(s) 1 > 2 sin x.

  The graphs intersect at about 0.524 or  and about 

2.618 or . The graph of y = 1 is above the other

graph when x < and when x > . Therefore, 1

> 2 sin x on 0 ≤ x <    or   < x < 2π.

51. 0 < 2 cos x –

SOLUTION:  Graph y = 2 cos x –  on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0 < 2 cos x – .

  The zeros of y = 2 cos x –  are about 0.785 or  and about 5.498 or  . The graph of y = 0 is

below the other graph when 0 ≤ x <  or   < x <

2π. Therefore, 0 < 2 cos x –  on 0 ≤ x <  or   < 

x < 2π.

52.  ≥ 

SOLUTION:  

Graph y = and y =  on [0, 2π). Use

the intersect feature under the CALC menu to

determine on what interval(s) ≥  .

  The graphs intersect at about 1.047 or  and about 

2.094 or . The graph of y = is below the

other graph when   ≤ x ≤  .

Therefore, ≥   on  ≤ x ≤  .

53.  ≤ tan x cot x

SOLUTION:  

Graph y = and y = tan x cot x on [0, 2π).

Use the intersect feature under the CALC menu to

determine on what interval(s) ≤ tan x cot

x.  

  The graphs intersect at about 3.142 or π. The graph of y = tan x cot x is at or below the other graph when 0 ≤ x < 2π.  

Therefore, ≤ tan x cot x on 0 ≤ x < 2π.

54. cos x ≤

SOLUTION:  

Graph y = cos x and y =  on [0, 2π). Use the

intersect feature under the CALC menu to

determine on what interval(s) cos x ≤ .

  The graphs intersect at about 2.618 or  and about

3.665 or . The graph of y = cos x is below the

other graph when  ≤ x ≤  .

 

Therefore, cos x ≤ on  ≤ x ≤  .

55.  sin x − 1 < 0

SOLUTION:  Graph y =  sin x − 1. Use the zero feature underthe CALC menu to determine on what interval(s)

 sin x − 1 < 0.

  The zeros of  sin x − 1 are about 0.785 or  and

about 2.356 or . The graph is below the x-axis

when 0 ≤ x <  or   < x < 2π.

Therefore,  sin x − 1 < 0 on 0 ≤ x <  or   < 

x < 2π.

56. REFRACTION When light travels from one medium to another it bends or refracts, as shown.  

Refraction is described by n2 sinθ1 = n1 sinθ2, wheren1 is the index of refraction of the medium the light

is entering, n2 is the index of refraction of the medium the light is exiting, θ1 is the angle of

incidence, and θ2 is the angle of refraction. a. Find θ2 for each material shown if the angle of incidence is 40º and the index of refraction for air is 1.00.

b. If the angle of incidence is doubled to 80 , will the resulting angles of refraction be twice as large asthose found in part a?

SOLUTION:  a. glass:

  ice:

  plastic:

  water:

  b. Find the angle of refraction for one of the mediums using an angle of incidence of 80 . glass:

θ2 ≈ 25 when θ1 = 40 andθ2 ≈ 40.4 when θ1 = 80 . Therefore, the angle of refraction is not twice as large.

57. ERROR ANALYSIS Vijay and Alicia are solving tan2 x – tan x +  =   tan x. Vijay thinks that

the solutions are x =  + n , x =  + n , x =

 + n , and x =  + n . Alicia thinks that the

solutions are x =  + n and x =  + n . Is

either of them correct? Explain your reasoning.

SOLUTION:  First, solve tan2 x – tan x +  =   tan x.

  On [0, 2π) tan x = 1 when x =  and x =  and 

tan x = when x =  and x = .

  Sample answer: Therefore, Vijay’s solutions are correct; however, they are not stated in the simplest

form. For example, his solutions of x =  + n and

x =  + n could simply be stated as x =  + 

n because when n = 1,  + nπ  is equivalent to

.

CHALLENGE  Solve each equation on the interval [0, 2π].

58. 16 sin5 x + 2 sin x = 12 sin3 x

SOLUTION:  

On [0, 2π) sin x = 0 when x = 0 and x = π, sin x =

when x =  and x = , sin x = when

x =  and x = , sin x =  – when x =  and 

x = , and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are 0, , , , , , , ,

, and .

59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3

SOLUTION:  

On [0, 2 ) sin x = 1 when x = , sin x  = –1 when

x = , sin x =  – when x =  and x = ,

and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are , , , , , and .

60. REASONING Are the solutions of csc x =  andcot2 x + 1 = 2 equivalent? If so, verify your answer algebraically. If not, explain your reasoning.

SOLUTION:  

 

 

The solutions to csc x =  are   and  . Using

a Pythagorean identity, cot2 x + 1 = 2 simplifies to csc2 x = 2 or csc x = ± . Therefore, there are

two additional solutions when csc x = – :  and

. So, the solutions of csc x =  and cot2 x + 1

= 2 are not equivalent.

OPEN ENDED Write a trigonometric equation that has each of the following solutions.

61. 

SOLUTION:  Sample answer: When sin x = 0, x = 0 and x = π.

When sin x = , x =  and x = . So, one

equation that has solutions of 0, π, , and  is 

= 0 or sin2 x –  = 0. This

can be rewritten as 2 sin2 x =  sin x.

62. 

SOLUTION:  

Sample answer: When cos x = , x =  and x =

. When cos x = – , x =  and x = .

So, one equation that has solutions of , , ,

and  is  = 0 or cos2 x

–  = 0. This can be rewritten as 4 cos2 x = 3.

63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying identities.

SOLUTION:  When solving an equation, you use properties of equality to manipulate each side of the equation to isolate a variable. For example:

  When verifying an identity, you transform an expression on one side of the identity into the expression on the other side through a series of algebraic steps.For example:

Verify = .  

Verify each identity.

64.  =

SOLUTION:  

65.  =

SOLUTION:  

66.  +  = 1

SOLUTION:  

+  

= cos2 θ + sin2 θ            Reciprocal Identities= 1                               Pythagorean Identity

Find the value of each expression using the given information.

67. tan ; sin θ = , tan  > 0

SOLUTION:  Use the Pythagorean Identity that involves sin θ.

Since tan  is positive and sin  is positive, cos θ

must be positive. So, .

eSolutions Manual - Powered by Cognero Page 7

5-3 Solving Trigonometric Equations

Page 8: 5-3 Solving Trigonometric Equations the interval , the only solution is when x = 1.34. 39. x log x + 5x cos x = –2 SOLUTION: On the interval , the solutions are when x = 1.84 and

Solve each equation for all values of x.1. 5 sin x + 2 = sin x

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is  + 2nπ,  + 2nπ, .

2. 5 =  sec 2 x + 3

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

3. 2 = 4 cos2 x + 1

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on

the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

4. 4 tan x – 7 = 3 tan x – 6

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

5. 9 + cot2 x = 12

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

6. 2 – 10 sec x = 4 – 9 sec x

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

7. 3 csc x = 2 csc x +

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

8. 11 =  3 csc2 x + 7

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are , , , and . Solutions

on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

9. 6 tan2 x – 2 = 4

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

10. 9 + sin2 x = 10

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the interval

(– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the solutions is

 + 2nπ,  + 2nπ, .

11. 7 cot x –   = 4 cot x

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

12. 7 cos x = 5 cos x +

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ,   + 2nπ, .

Find all solutions of each equation on [0, 2 ).13. sin4 x + 2 sin2 x − 3 = 0

SOLUTION:  

On the interval [0, 2π),  when x =  and 

 when x = . Since  is not a real 

number, the equation  yields no additional solutions.

14. –2 sin x = –sin x cos x

SOLUTION:  

The equation cos x = 2 has no real solutions since the maximum value the cosine function can obtain is 1. On the interval [0, 2π), the equation  has solutions 0 and π.

15. 4 cot x = cot x sin2 x

SOLUTION:  

The equations sin x = 2 and sin x = –2 have no real solutions. On the interval [0, 2π), the equation cot x =

0 has solutions  and  .

16. csc2 x – csc x + 9 = 11

SOLUTION:  

On the interval [0, 2π), the equation csc x = –1 has a

solution of  and the equation csc x = 2 has

solutions of    and  .

17. cos3 x + cos2 x – cos x = 1

SOLUTION:  

On the interval [0, 2π), the equation cos x = 1 has a solution of 0 and the equation cos x = –1 has a solution of  π.

18. 2 sin2 x = sin x + 1

SOLUTION:  

On the interval [0, 2π), the equation sin x = 1 has a

solution of  and the equation sin x =  has  

solutions of    and  .

19. TENNIS A tennis ball leaves a racquet and heads toward a net 40 feet away. The height of the net is the same height as the initial height of the tennis ball. 

a. If the ball is hit at 50 feet per second, neglecting air resistance, use  to find the interval of possible angles of the ball needed to clear the net. b. Find  if the initial velocity remained the same but the distance to the net was 50 feet.

SOLUTION:  a.

The interval is [15.4°, 74.6°].   b.

If the distance to the net is 50 feet, then the angle would be 19.9° or 70.1°.

20. SKIING In the Olympics aerial skiing competition, skiers speed down a slope that launches them into the air, as shown. The maximum height a skier can

reach is given by hpeak =  , where g is

the acceleration due to gravity or 9.8 meters per second squared.

a. If a skier obtains a height of 5 meters above the end of the ramp, what was the skier’s initial speed? b. Use your answer from part a to determine how long it took the skier to reach the maximum height if tpeak = .

SOLUTION:  a.

The skier’s initial speed was 11.67 meters per second.   b.

  The time it took the skier to reach the maximum height was about 1.0 seconds.

Find all solutions of each equation on the interval [0, 2 ).

21. 1 = cot 2 x + csc x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

22. sec x = tan x + 1

SOLUTION:  

   

  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is 0.

23. tan 2 x = 1 – sec x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are 0,

, and .

24. csc x + cot x = 1

SOLUTION:  

 

x =  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is .

25. 2 – 2 cos2 x = sin x + 1

SOLUTION:  

Check

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

26. cos x – 4 = sin x – 4

SOLUTION:   

 

Therefore, on the interval [0, 2π) the only valid

solutions are and .

27. 3 sin x = 3 – 3 cos x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the only valid

solutions are and 0.

28. cot2 x csc2 x – cot2 x = 9

SOLUTION:  

is undefined. The solutions of  are 

  Check

 

 

 

On the interval , there are no real values of x for which , but  for 

29. sec2 x – 1 + tan x –  tan x =

SOLUTION:  

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

30. sec2 x tan2 x + 3 sec2 x – 2 tan2 x = 3

SOLUTION:  

 

The square of any real number must be greater than or equal to zero so sec2 x = –1 has no solutions. Therefore, on the interval [0, 2π) the only solutions are 0 and π.

31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive power PR of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical components, PH and PV.

Using the equations above, determine for what values of  PV and PH are equivalent.

SOLUTION:  

The sine and cosine have the same values in the

interval [0, 2π) at  and  .

Therefore, the components will be equivalent when

.

Find all solutions of each equation on the interval [0, 2 ).

32.  + cos x = 2

SOLUTION:  

On the interval [0, 2π), cos x =  when x =  and 

when x = .

33.  +  = –4

SOLUTION:  

On the interval [0, 2π),cos x = – when x =

 and when x = .

34.  +  = cos x

SOLUTION:  

On the interval [0, 2π) there are no values of x for

which sin x = –2 and the sin x = 1 only when x = .

However, tan  is undefined, so there are no 

solutions for the original equation.

35. cot x cos x + 1 =  + 

SOLUTION:  

On ,  when x =  and when x =

.

GRAPHING CALCULATOR Solve each equation on the interval [0, 2 ) by graphing. Round to the nearest hundredth.

36. 3 cos 2x = ex + 1

SOLUTION:  

On the interval , the only solution is when x = 0.33.

37. sin x + cos x = 3x

SOLUTION:  

On the interval , the only solution is when x = 0.41.

38. x2 = 2 cos x + x

SOLUTION:  

On the interval , the only solution is when x = 1.34.

39. x log x + 5x cos x = –2

SOLUTION:  

On the interval , the solutions are when x = 1.84 and when x = 4.49.

40. METEOROLOGY The average daily temperature in degrees Fahrenheit for a city can be modeled by t

= 8.05 cos + 66.95, where x is a function of

time, x = 1 represents January 15, x = 2 represents February 15, and so on. a. Use a graphing calculator to estimate the temperature on January 31. b. Approximate the number of months that the average daily temperature is greater than 70 throughout the entire month. c. Estimate the highest temperature of the year and the month in which it occurs.

SOLUTION:  a. Use a graphing calculator to graph t = 8.05

cos + 66.95. Because x = 1 represents

January 15, x = 1.5 represents January 31. Use the TRACE feature to determine the temperature whenx = 1.5.

Therefore, the temperature on January 31 is about 61.3º.   b. Graph y = 70. Then use the intersect feature under the CALC menu to determine the interval on

which 8.05 cos + 66.95 > 70.

So, 8.05 cos + 66.95 > 70 on the interval

3.74 < x < 8.26. Because x = 3.74 corresponds to about the middle of April, the temperature does not exceed 70º for the entire month until May. Therefore, the average daily temperature is greater than 70º for the entire month during the months of May, June, and July. Therefore, the average temperature was higher than 70º for three months.   c. Use the maximum feature under the CALC menu to find the maximum value of the function on [0, 12].

The maximum value is 75 when x ≈ 6. Therefore, the maximum average monthly temperature is 75º during the month of June.

Find the x-intercepts of each graph on the interval [0, 2 ).

41. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) cos x = 0 when x =  and x

= .

42. 

 

SOLUTION:  Let y = 0 and solve for x.

  On the interval [0, 2π) sin x = 0 when x = 0 and x = π.

43. 

SOLUTION:  Let y = 0 and solve for x.

 

On the interval [0, 2π) cot x = 1 when x =  and x

= .

44. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) tan x = 0 when x = 0 and x =

π, tan x = –1 when x =  and x = , and tan x

= 1 when x =  and x = .

Find all solutions of each equation on the interval [0, 4 ).

45. 4 tan x = 2 sec2 x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

46. 2 sin2 x + 1 = –3 sin x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, , , and .

47. csc x cot2 x = csc x

SOLUTION:   

On the interval [0, 4π) the solutions are , ,

, , , , , and .

48. sec x + 5 = 2 sec x + 3

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

49. GEOMETRY Consider the circle below.

a. The length s of  is given by s = r(2 ) where 0≤   ≤  . When s = 18 and AB = 14, the radius is r

= . Use a graphing calculator to find the

measure of 2 in radians. b. The area of the shaded region is given by A =

. Use a graphing calculator to find the

radian measure of θ  if the radius is 5 inches and the area is 36 square inches. Round to the nearest hundredth.

SOLUTION:  a. Rewrite the arclength formula using s = 18 and r

= .

On the graphing calculator, find the intersection of Y1 = 18sinθ and Y2 = 14θ.

The value of 2θ = 2(1.1968) or about 2.39 radians.     b. First, substitute into the given area formula and rearrange it.

  Using a graphing calculator, find the intersection of Y1 = 2.88 and Y2 = θ – sin θ.

When the area is 36 square inches and the radius is 5 inches, then the measure of θ is 3.01 radians.

Solve each inequality on the interval [0, 2 ).50. 1 > 2 sin x

SOLUTION:  Graph y = 1 and y = 2 sin x on [0, 2π). Use the intersect feature under the CALC menu to determine on what interval(s) 1 > 2 sin x.

  The graphs intersect at about 0.524 or  and about 

2.618 or . The graph of y = 1 is above the other

graph when x < and when x > . Therefore, 1

> 2 sin x on 0 ≤ x <    or   < x < 2π.

51. 0 < 2 cos x –

SOLUTION:  Graph y = 2 cos x –  on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0 < 2 cos x – .

  The zeros of y = 2 cos x –  are about 0.785 or  and about 5.498 or  . The graph of y = 0 is

below the other graph when 0 ≤ x <  or   < x <

2π. Therefore, 0 < 2 cos x –  on 0 ≤ x <  or   < 

x < 2π.

52.  ≥ 

SOLUTION:  

Graph y = and y =  on [0, 2π). Use

the intersect feature under the CALC menu to

determine on what interval(s) ≥  .

  The graphs intersect at about 1.047 or  and about 

2.094 or . The graph of y = is below the

other graph when   ≤ x ≤  .

Therefore, ≥   on  ≤ x ≤  .

53.  ≤ tan x cot x

SOLUTION:  

Graph y = and y = tan x cot x on [0, 2π).

Use the intersect feature under the CALC menu to

determine on what interval(s) ≤ tan x cot

x.  

  The graphs intersect at about 3.142 or π. The graph of y = tan x cot x is at or below the other graph when 0 ≤ x < 2π.  

Therefore, ≤ tan x cot x on 0 ≤ x < 2π.

54. cos x ≤

SOLUTION:  

Graph y = cos x and y =  on [0, 2π). Use the

intersect feature under the CALC menu to

determine on what interval(s) cos x ≤ .

  The graphs intersect at about 2.618 or  and about

3.665 or . The graph of y = cos x is below the

other graph when  ≤ x ≤  .

 

Therefore, cos x ≤ on  ≤ x ≤  .

55.  sin x − 1 < 0

SOLUTION:  Graph y =  sin x − 1. Use the zero feature underthe CALC menu to determine on what interval(s)

 sin x − 1 < 0.

  The zeros of  sin x − 1 are about 0.785 or  and

about 2.356 or . The graph is below the x-axis

when 0 ≤ x <  or   < x < 2π.

Therefore,  sin x − 1 < 0 on 0 ≤ x <  or   < 

x < 2π.

56. REFRACTION When light travels from one medium to another it bends or refracts, as shown.  

Refraction is described by n2 sinθ1 = n1 sinθ2, wheren1 is the index of refraction of the medium the light

is entering, n2 is the index of refraction of the medium the light is exiting, θ1 is the angle of

incidence, and θ2 is the angle of refraction. a. Find θ2 for each material shown if the angle of incidence is 40º and the index of refraction for air is 1.00.

b. If the angle of incidence is doubled to 80 , will the resulting angles of refraction be twice as large asthose found in part a?

SOLUTION:  a. glass:

  ice:

  plastic:

  water:

  b. Find the angle of refraction for one of the mediums using an angle of incidence of 80 . glass:

θ2 ≈ 25 when θ1 = 40 andθ2 ≈ 40.4 when θ1 = 80 . Therefore, the angle of refraction is not twice as large.

57. ERROR ANALYSIS Vijay and Alicia are solving tan2 x – tan x +  =   tan x. Vijay thinks that

the solutions are x =  + n , x =  + n , x =

 + n , and x =  + n . Alicia thinks that the

solutions are x =  + n and x =  + n . Is

either of them correct? Explain your reasoning.

SOLUTION:  First, solve tan2 x – tan x +  =   tan x.

  On [0, 2π) tan x = 1 when x =  and x =  and 

tan x = when x =  and x = .

  Sample answer: Therefore, Vijay’s solutions are correct; however, they are not stated in the simplest

form. For example, his solutions of x =  + n and

x =  + n could simply be stated as x =  + 

n because when n = 1,  + nπ  is equivalent to

.

CHALLENGE  Solve each equation on the interval [0, 2π].

58. 16 sin5 x + 2 sin x = 12 sin3 x

SOLUTION:  

On [0, 2π) sin x = 0 when x = 0 and x = π, sin x =

when x =  and x = , sin x = when

x =  and x = , sin x =  – when x =  and 

x = , and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are 0, , , , , , , ,

, and .

59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3

SOLUTION:  

On [0, 2 ) sin x = 1 when x = , sin x  = –1 when

x = , sin x =  – when x =  and x = ,

and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are , , , , , and .

60. REASONING Are the solutions of csc x =  andcot2 x + 1 = 2 equivalent? If so, verify your answer algebraically. If not, explain your reasoning.

SOLUTION:  

 

 

The solutions to csc x =  are   and  . Using

a Pythagorean identity, cot2 x + 1 = 2 simplifies to csc2 x = 2 or csc x = ± . Therefore, there are

two additional solutions when csc x = – :  and

. So, the solutions of csc x =  and cot2 x + 1

= 2 are not equivalent.

OPEN ENDED Write a trigonometric equation that has each of the following solutions.

61. 

SOLUTION:  Sample answer: When sin x = 0, x = 0 and x = π.

When sin x = , x =  and x = . So, one

equation that has solutions of 0, π, , and  is 

= 0 or sin2 x –  = 0. This

can be rewritten as 2 sin2 x =  sin x.

62. 

SOLUTION:  

Sample answer: When cos x = , x =  and x =

. When cos x = – , x =  and x = .

So, one equation that has solutions of , , ,

and  is  = 0 or cos2 x

–  = 0. This can be rewritten as 4 cos2 x = 3.

63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying identities.

SOLUTION:  When solving an equation, you use properties of equality to manipulate each side of the equation to isolate a variable. For example:

  When verifying an identity, you transform an expression on one side of the identity into the expression on the other side through a series of algebraic steps.For example:

Verify = .  

Verify each identity.

64.  =

SOLUTION:  

65.  =

SOLUTION:  

66.  +  = 1

SOLUTION:  

+  

= cos2 θ + sin2 θ            Reciprocal Identities= 1                               Pythagorean Identity

Find the value of each expression using the given information.

67. tan ; sin θ = , tan  > 0

SOLUTION:  Use the Pythagorean Identity that involves sin θ.

Since tan  is positive and sin  is positive, cos θ

must be positive. So, .

Solve each equation for all values of x.1. 5 sin x + 2 = sin x

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is  + 2nπ,  + 2nπ, .

2. 5 =  sec 2 x + 3

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

3. 2 = 4 cos2 x + 1

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are , , , and . Solutions on

the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

4. 4 tan x – 7 = 3 tan x – 6

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

5. 9 + cot2 x = 12

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

6. 2 – 10 sec x = 4 – 9 sec x

SOLUTION:  

The period of secant is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

7. 3 csc x = 2 csc x +

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, .

8. 11 =  3 csc2 x + 7

SOLUTION:  

The period of cosecant is 2π, so you only need to find solutions on the interval . The solutions

on this interval are , , , and . Solutions

on the interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ, + 2nπ, + 2nπ, +

2nπ, .

9. 6 tan2 x – 2 = 4

SOLUTION:  

The period of tangent is π, so you only need to find solutions on the interval . The solutions on this

interval are and . Solutions on the interval (–

, ), are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, + nπ, .

10. 9 + sin2 x = 10

SOLUTION:  

The period of sine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are  and  . Solutions on the interval

(– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the solutions is

 + 2nπ,  + 2nπ, .

11. 7 cot x –   = 4 cot x

SOLUTION:  

The period of cotangent is π, so you only need to findsolutions on the interval . The only solution on

this interval is . Solutions on the interval (– , ),

are found by adding integer multiples of π.  

Therefore, the general form of the solutions is +

nπ, .

12. 7 cos x = 5 cos x +

SOLUTION:  

The period of cosine is 2π, so you only need to find solutions on the interval . The solutions on

this interval are and . Solutions on the

interval (– , ), are found by adding integer multiples of 2π.  Therefore, the general form of the

solutions is + 2nπ,   + 2nπ, .

Find all solutions of each equation on [0, 2 ).13. sin4 x + 2 sin2 x − 3 = 0

SOLUTION:  

On the interval [0, 2π),  when x =  and 

 when x = . Since  is not a real 

number, the equation  yields no additional solutions.

14. –2 sin x = –sin x cos x

SOLUTION:  

The equation cos x = 2 has no real solutions since the maximum value the cosine function can obtain is 1. On the interval [0, 2π), the equation  has solutions 0 and π.

15. 4 cot x = cot x sin2 x

SOLUTION:  

The equations sin x = 2 and sin x = –2 have no real solutions. On the interval [0, 2π), the equation cot x =

0 has solutions  and  .

16. csc2 x – csc x + 9 = 11

SOLUTION:  

On the interval [0, 2π), the equation csc x = –1 has a

solution of  and the equation csc x = 2 has

solutions of    and  .

17. cos3 x + cos2 x – cos x = 1

SOLUTION:  

On the interval [0, 2π), the equation cos x = 1 has a solution of 0 and the equation cos x = –1 has a solution of  π.

18. 2 sin2 x = sin x + 1

SOLUTION:  

On the interval [0, 2π), the equation sin x = 1 has a

solution of  and the equation sin x =  has  

solutions of    and  .

19. TENNIS A tennis ball leaves a racquet and heads toward a net 40 feet away. The height of the net is the same height as the initial height of the tennis ball. 

a. If the ball is hit at 50 feet per second, neglecting air resistance, use  to find the interval of possible angles of the ball needed to clear the net. b. Find  if the initial velocity remained the same but the distance to the net was 50 feet.

SOLUTION:  a.

The interval is [15.4°, 74.6°].   b.

If the distance to the net is 50 feet, then the angle would be 19.9° or 70.1°.

20. SKIING In the Olympics aerial skiing competition, skiers speed down a slope that launches them into the air, as shown. The maximum height a skier can

reach is given by hpeak =  , where g is

the acceleration due to gravity or 9.8 meters per second squared.

a. If a skier obtains a height of 5 meters above the end of the ramp, what was the skier’s initial speed? b. Use your answer from part a to determine how long it took the skier to reach the maximum height if tpeak = .

SOLUTION:  a.

The skier’s initial speed was 11.67 meters per second.   b.

  The time it took the skier to reach the maximum height was about 1.0 seconds.

Find all solutions of each equation on the interval [0, 2 ).

21. 1 = cot 2 x + csc x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

22. sec x = tan x + 1

SOLUTION:  

   

  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is 0.

23. tan 2 x = 1 – sec x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the solutions are 0,

, and .

24. csc x + cot x = 1

SOLUTION:  

 

x =  is an extraneous solution. Therefore, on the

interval [0, 2π) the only solution is .

25. 2 – 2 cos2 x = sin x + 1

SOLUTION:  

Check

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

26. cos x – 4 = sin x – 4

SOLUTION:   

 

Therefore, on the interval [0, 2π) the only valid

solutions are and .

27. 3 sin x = 3 – 3 cos x

SOLUTION:  

 

 

 

Therefore, on the interval [0, 2π) the only valid

solutions are and 0.

28. cot2 x csc2 x – cot2 x = 9

SOLUTION:  

is undefined. The solutions of  are 

  Check

 

 

 

On the interval , there are no real values of x for which , but  for 

29. sec2 x – 1 + tan x –  tan x =

SOLUTION:  

 

 

Therefore, on the interval [0, 2π) the solutions are

, , and .

30. sec2 x tan2 x + 3 sec2 x – 2 tan2 x = 3

SOLUTION:  

 

The square of any real number must be greater than or equal to zero so sec2 x = –1 has no solutions. Therefore, on the interval [0, 2π) the only solutions are 0 and π.

31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive power PR of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical components, PH and PV.

Using the equations above, determine for what values of  PV and PH are equivalent.

SOLUTION:  

The sine and cosine have the same values in the

interval [0, 2π) at  and  .

Therefore, the components will be equivalent when

.

Find all solutions of each equation on the interval [0, 2 ).

32.  + cos x = 2

SOLUTION:  

On the interval [0, 2π), cos x =  when x =  and 

when x = .

33.  +  = –4

SOLUTION:  

On the interval [0, 2π),cos x = – when x =

 and when x = .

34.  +  = cos x

SOLUTION:  

On the interval [0, 2π) there are no values of x for

which sin x = –2 and the sin x = 1 only when x = .

However, tan  is undefined, so there are no 

solutions for the original equation.

35. cot x cos x + 1 =  + 

SOLUTION:  

On ,  when x =  and when x =

.

GRAPHING CALCULATOR Solve each equation on the interval [0, 2 ) by graphing. Round to the nearest hundredth.

36. 3 cos 2x = ex + 1

SOLUTION:  

On the interval , the only solution is when x = 0.33.

37. sin x + cos x = 3x

SOLUTION:  

On the interval , the only solution is when x = 0.41.

38. x2 = 2 cos x + x

SOLUTION:  

On the interval , the only solution is when x = 1.34.

39. x log x + 5x cos x = –2

SOLUTION:  

On the interval , the solutions are when x = 1.84 and when x = 4.49.

40. METEOROLOGY The average daily temperature in degrees Fahrenheit for a city can be modeled by t

= 8.05 cos + 66.95, where x is a function of

time, x = 1 represents January 15, x = 2 represents February 15, and so on. a. Use a graphing calculator to estimate the temperature on January 31. b. Approximate the number of months that the average daily temperature is greater than 70 throughout the entire month. c. Estimate the highest temperature of the year and the month in which it occurs.

SOLUTION:  a. Use a graphing calculator to graph t = 8.05

cos + 66.95. Because x = 1 represents

January 15, x = 1.5 represents January 31. Use the TRACE feature to determine the temperature whenx = 1.5.

Therefore, the temperature on January 31 is about 61.3º.   b. Graph y = 70. Then use the intersect feature under the CALC menu to determine the interval on

which 8.05 cos + 66.95 > 70.

So, 8.05 cos + 66.95 > 70 on the interval

3.74 < x < 8.26. Because x = 3.74 corresponds to about the middle of April, the temperature does not exceed 70º for the entire month until May. Therefore, the average daily temperature is greater than 70º for the entire month during the months of May, June, and July. Therefore, the average temperature was higher than 70º for three months.   c. Use the maximum feature under the CALC menu to find the maximum value of the function on [0, 12].

The maximum value is 75 when x ≈ 6. Therefore, the maximum average monthly temperature is 75º during the month of June.

Find the x-intercepts of each graph on the interval [0, 2 ).

41. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) cos x = 0 when x =  and x

= .

42. 

 

SOLUTION:  Let y = 0 and solve for x.

  On the interval [0, 2π) sin x = 0 when x = 0 and x = π.

43. 

SOLUTION:  Let y = 0 and solve for x.

 

On the interval [0, 2π) cot x = 1 when x =  and x

= .

44. 

SOLUTION:  Let y = 0 and solve for x.

On the interval [0, 2π) tan x = 0 when x = 0 and x =

π, tan x = –1 when x =  and x = , and tan x

= 1 when x =  and x = .

Find all solutions of each equation on the interval [0, 4 ).

45. 4 tan x = 2 sec2 x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

46. 2 sin2 x + 1 = –3 sin x

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, , , and .

47. csc x cot2 x = csc x

SOLUTION:   

On the interval [0, 4π) the solutions are , ,

, , , , , and .

48. sec x + 5 = 2 sec x + 3

SOLUTION:  

On the interval [0, 4π) the solutions are , ,

, and .

49. GEOMETRY Consider the circle below.

a. The length s of  is given by s = r(2 ) where 0≤   ≤  . When s = 18 and AB = 14, the radius is r

= . Use a graphing calculator to find the

measure of 2 in radians. b. The area of the shaded region is given by A =

. Use a graphing calculator to find the

radian measure of θ  if the radius is 5 inches and the area is 36 square inches. Round to the nearest hundredth.

SOLUTION:  a. Rewrite the arclength formula using s = 18 and r

= .

On the graphing calculator, find the intersection of Y1 = 18sinθ and Y2 = 14θ.

The value of 2θ = 2(1.1968) or about 2.39 radians.     b. First, substitute into the given area formula and rearrange it.

  Using a graphing calculator, find the intersection of Y1 = 2.88 and Y2 = θ – sin θ.

When the area is 36 square inches and the radius is 5 inches, then the measure of θ is 3.01 radians.

Solve each inequality on the interval [0, 2 ).50. 1 > 2 sin x

SOLUTION:  Graph y = 1 and y = 2 sin x on [0, 2π). Use the intersect feature under the CALC menu to determine on what interval(s) 1 > 2 sin x.

  The graphs intersect at about 0.524 or  and about 

2.618 or . The graph of y = 1 is above the other

graph when x < and when x > . Therefore, 1

> 2 sin x on 0 ≤ x <    or   < x < 2π.

51. 0 < 2 cos x –

SOLUTION:  Graph y = 2 cos x –  on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0 < 2 cos x – .

  The zeros of y = 2 cos x –  are about 0.785 or  and about 5.498 or  . The graph of y = 0 is

below the other graph when 0 ≤ x <  or   < x <

2π. Therefore, 0 < 2 cos x –  on 0 ≤ x <  or   < 

x < 2π.

52.  ≥ 

SOLUTION:  

Graph y = and y =  on [0, 2π). Use

the intersect feature under the CALC menu to

determine on what interval(s) ≥  .

  The graphs intersect at about 1.047 or  and about 

2.094 or . The graph of y = is below the

other graph when   ≤ x ≤  .

Therefore, ≥   on  ≤ x ≤  .

53.  ≤ tan x cot x

SOLUTION:  

Graph y = and y = tan x cot x on [0, 2π).

Use the intersect feature under the CALC menu to

determine on what interval(s) ≤ tan x cot

x.  

  The graphs intersect at about 3.142 or π. The graph of y = tan x cot x is at or below the other graph when 0 ≤ x < 2π.  

Therefore, ≤ tan x cot x on 0 ≤ x < 2π.

54. cos x ≤

SOLUTION:  

Graph y = cos x and y =  on [0, 2π). Use the

intersect feature under the CALC menu to

determine on what interval(s) cos x ≤ .

  The graphs intersect at about 2.618 or  and about

3.665 or . The graph of y = cos x is below the

other graph when  ≤ x ≤  .

 

Therefore, cos x ≤ on  ≤ x ≤  .

55.  sin x − 1 < 0

SOLUTION:  Graph y =  sin x − 1. Use the zero feature underthe CALC menu to determine on what interval(s)

 sin x − 1 < 0.

  The zeros of  sin x − 1 are about 0.785 or  and

about 2.356 or . The graph is below the x-axis

when 0 ≤ x <  or   < x < 2π.

Therefore,  sin x − 1 < 0 on 0 ≤ x <  or   < 

x < 2π.

56. REFRACTION When light travels from one medium to another it bends or refracts, as shown.  

Refraction is described by n2 sinθ1 = n1 sinθ2, wheren1 is the index of refraction of the medium the light

is entering, n2 is the index of refraction of the medium the light is exiting, θ1 is the angle of

incidence, and θ2 is the angle of refraction. a. Find θ2 for each material shown if the angle of incidence is 40º and the index of refraction for air is 1.00.

b. If the angle of incidence is doubled to 80 , will the resulting angles of refraction be twice as large asthose found in part a?

SOLUTION:  a. glass:

  ice:

  plastic:

  water:

  b. Find the angle of refraction for one of the mediums using an angle of incidence of 80 . glass:

θ2 ≈ 25 when θ1 = 40 andθ2 ≈ 40.4 when θ1 = 80 . Therefore, the angle of refraction is not twice as large.

57. ERROR ANALYSIS Vijay and Alicia are solving tan2 x – tan x +  =   tan x. Vijay thinks that

the solutions are x =  + n , x =  + n , x =

 + n , and x =  + n . Alicia thinks that the

solutions are x =  + n and x =  + n . Is

either of them correct? Explain your reasoning.

SOLUTION:  First, solve tan2 x – tan x +  =   tan x.

  On [0, 2π) tan x = 1 when x =  and x =  and 

tan x = when x =  and x = .

  Sample answer: Therefore, Vijay’s solutions are correct; however, they are not stated in the simplest

form. For example, his solutions of x =  + n and

x =  + n could simply be stated as x =  + 

n because when n = 1,  + nπ  is equivalent to

.

CHALLENGE  Solve each equation on the interval [0, 2π].

58. 16 sin5 x + 2 sin x = 12 sin3 x

SOLUTION:  

On [0, 2π) sin x = 0 when x = 0 and x = π, sin x =

when x =  and x = , sin x = when

x =  and x = , sin x =  – when x =  and 

x = , and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are 0, , , , , , , ,

, and .

59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3

SOLUTION:  

On [0, 2 ) sin x = 1 when x = , sin x  = –1 when

x = , sin x =  – when x =  and x = ,

and sin x =  when x =  and x = .

  Therefore, after checking for extraneous solutions,

the solutions are , , , , , and .

60. REASONING Are the solutions of csc x =  andcot2 x + 1 = 2 equivalent? If so, verify your answer algebraically. If not, explain your reasoning.

SOLUTION:  

 

 

The solutions to csc x =  are   and  . Using

a Pythagorean identity, cot2 x + 1 = 2 simplifies to csc2 x = 2 or csc x = ± . Therefore, there are

two additional solutions when csc x = – :  and

. So, the solutions of csc x =  and cot2 x + 1

= 2 are not equivalent.

OPEN ENDED Write a trigonometric equation that has each of the following solutions.

61. 

SOLUTION:  Sample answer: When sin x = 0, x = 0 and x = π.

When sin x = , x =  and x = . So, one

equation that has solutions of 0, π, , and  is 

= 0 or sin2 x –  = 0. This

can be rewritten as 2 sin2 x =  sin x.

62. 

SOLUTION:  

Sample answer: When cos x = , x =  and x =

. When cos x = – , x =  and x = .

So, one equation that has solutions of , , ,

and  is  = 0 or cos2 x

–  = 0. This can be rewritten as 4 cos2 x = 3.

63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying identities.

SOLUTION:  When solving an equation, you use properties of equality to manipulate each side of the equation to isolate a variable. For example:

  When verifying an identity, you transform an expression on one side of the identity into the expression on the other side through a series of algebraic steps.For example:

Verify = .  

Verify each identity.

64.  =

SOLUTION:  

65.  =

SOLUTION:  

66.  +  = 1

SOLUTION:  

+  

= cos2 θ + sin2 θ            Reciprocal Identities= 1                               Pythagorean Identity

Find the value of each expression using the given information.

67. tan ; sin θ = , tan  > 0

SOLUTION:  Use the Pythagorean Identity that involves sin θ.

Since tan  is positive and sin  is positive, cos θ

must be positive. So, .

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5-3 Solving Trigonometric Equations


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