Trigonometry

Sec. 03 notes

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Solving Trig Equations: The Others

Main Idea

In the last sections, we learned to solve very basic trig equations as well as slight variations of these basic ones. Inthis section, we show techniques which can be used to turn other equations into basic ones. The use of identitieswill be essential. The general strategy is to get all expressions on one side of the equation and 0 on the other side.Then we try to factor the expressions into simpler terms so that we may use the Zero Factor Theorem. This is,generally speaking, easier said than done. Yet for our class, in may be appropriate so propose mostly equationswhich do in fact factor in a relatively easy way, so long as we have full command and use of the famous identities.

The following suggestions may often prove helpful.

Solving Trig Eqs: SomeStrategies

• Tweak to turn into ez ones.

• Turn everything into sines or cosines.

• Reduce angles so they are all equal.

• Use the famous identities

• Use conjugates

• See if you can factor as a quadratic equation

• Careful not to mult/or divide by zero

• Check answers when finished.

Example:

Solve24sin2(x) + − 38sin(x) + 15 = 0

24sin2(x) + − 38sin(x) + 15 = 0 (given)[

4sin(x) + − 3]

·[

6sin(x) + − 5]

= 0 (factor)

4sin(x) + − 3 = 0 OR 6sin(x) + − 5 = 0 (Zero Fact Thm)

sin(x) =3

4OR sin(x) =

5

6(algebra)

Solve

sin (x) =3

4

Solution:

xk ≈ 48.59◦ + k360◦ for k ∈ Z

ORxk ≈ 131.41◦ + k360◦ for k ∈ Z

Solve

sin (x) =5

6

Solution:

xk ≈ 56.443◦ + k360◦ for k ∈ Z

ORxk ≈ 123.557◦ + k360◦ for k ∈ Z

Example:

Solve6cos2(x) + − 13cos(x) + 5 = 0

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6cos2(x) + − 13cos(x) + 5 = 0 (given)[

2cos(x) + − 1]

·[

3cos(x) + − 5]

= 0 (factor)

2cos(x) + − 1 = 0 OR 3cos(x) + − 5 = 0 (Zero Fact Thm)

cos(x) =1

2OR cos(x) =

5

3(algebra)

Solve

cos (x) =1

2

Solution:

xk ≈ 60◦ + k360◦ for k ∈ Z

ORxk ≈ 300◦ + k360◦ for k ∈ Z

Solve

cos (x) =5

3

no real solution for x

Example:

Solve2sin2(x) + − 5sin(x) + − 3 = 0

2sin2(x) + − 5sin(x) + − 3 = 0 (given)[

2sin(x) + 1]

·[

1sin(x) + − 3]

= 0 (factor)

2sin(x) + 1 = 0 OR 1sin(x) + − 3 = 0 (Zero Fact Thm)

sin(x) = −1

2OR sin(x) = 3 (algebra)

Solve

sin (x) = −1

2

Solution:

xk ≈ −30◦ + k360◦ for k ∈ Z

ORxk ≈ 210◦ + k360◦ for k ∈ Z

Solve

sin (x) = 3

no real solution for x

c©2007-2009 MathHands.commathhands pg. 2

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Solving Trig Equations: The Others

1. Solve3sin2(x) + sin(x) = 0

Solution:

3sin2(x) + 1sin(x) = 0 (given)

sin(x)[

3sin(x) + 1]

= 0 (factor)

sin(x) = 0 OR 3sin(x) + 1 = 0 (Zero Fact Thm)

sin(x) = 0 OR sin(x) = −1

3(algebra)

Solvesin (x) = 0

Solution:

xk = 0◦ + k180◦ for k ∈ Z

Solve

sin (x) = −1

3

Solution:

xk ≈ −19.471◦ + k360◦ for k ∈ Z

OR

xk ≈ 199.471◦ + k360◦ for k ∈ Z

2. Solve2sin2(x) + 5sin(x) = 0

Solution:

2sin2(x) + 5sin(x) = 0 (given)

sin(x)[

2sin(x) + 5]

= 0 (factor)

sin(x) = 0 OR 2sin(x) + 5 = 0 (Zero Fact Thm)

sin(x) = 0 OR sin(x) = −5

2(algebra)

Solve

sin (x) = 0

Solution:

xk = 0◦ + k180◦ for k ∈ Z

Solve

sin (x) = −5

2

no real solution for x

3. Solve5cos2(x) + cos(x) = 0

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Solution:

5cos2(x) + 1cos(x) = 0 (given)

cos(x)[

5cos(x) + 1]

= 0 (factor)

cos(x) = 0 OR 5cos(x) + 1 = 0 (Zero Fact Thm)

cos(x) = 0 OR cos(x) = −1

5(algebra)

Solvecos (x) = 0

Solution:

xk = 90◦ + k180◦ for k ∈ Z

Solve

cos (x) = −1

5

Solution:

xk ≈ 101.537◦ + k360◦ for k ∈ Z

OR

xk ≈ 258.463◦ + k360◦ for k ∈ Z

4. Solve6cos2(x) + 2cos(x) = 0

Solution:

6cos2(x) + 2cos(x) = 0 (given)

cos(x)[

6cos(x) + 2]

= 0 (factor)

cos(x) = 0 OR 6cos(x) + 2 = 0 (Zero Fact Thm)

cos(x) = 0 OR cos(x) = −1

3(algebra)

Solvecos (x) = 0

Solution:

xk = 90◦ + k180◦ for k ∈ Z

Solve

cos (x) = −1

3

Solution:

xk ≈ 109.471◦ + k360◦ for k ∈ Z

OR

xk ≈ 250.529◦ + k360◦ for k ∈ Z

5. Solve8sin2(x) + − 2sin(x) + − 3 = 0

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Solution:

8sin2(x) + − 2sin(x) + − 3 = 0 (given)[

2sin(x) + 1]

·[

4sin(x) + − 3]

= 0 (factor)

2sin(x) + 1 = 0 OR 4sin(x) + − 3 = 0 (Zero Fact Thm)

sin(x) = −1

2OR sin(x) =

3

4(algebra)

Solve

sin (x) = −1

2

Solution:

xk ≈ −30◦ + k360◦ for k ∈ Z

ORxk ≈ 210◦ + k360◦ for k ∈ Z

Solve

sin (x) =3

4

Solution:

xk ≈ 48.59◦ + k360◦ for k ∈ Z

ORxk ≈ 131.41◦ + k360◦ for k ∈ Z

6. Solve4sin2(x) + 0sin(x) + − 1 = 0

Solution:

4sin2(x) + 0sin(x) + − 1 = 0 (given)[

2sin(x) + 1]

·[

2sin(x) + − 1]

= 0 (factor)

2sin(x) + 1 = 0 OR 2sin(x) + − 1 = 0 (Zero Fact Thm)

sin(x) = −1

2OR sin(x) =

1

2(algebra)

Solve

sin (x) = −1

2

Solution:

xk ≈ −30◦ + k360◦ for k ∈ Z

ORxk ≈ 210◦ + k360◦ for k ∈ Z

Solve

sin (x) =1

2

Solution:

xk ≈ 30◦ + k360◦ for k ∈ Z

ORxk ≈ 150◦ + k360◦ for k ∈ Z

7. Solve4cos2(x) + 0cos(x) + − 1 = 0

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Solution:

4cos2(x) + 0cos(x) + − 1 = 0 (given)[

2cos(x) + 1]

·[

2cos(x) + − 1]

= 0 (factor)

2cos(x) + 1 = 0 OR 2cos(x) + − 1 = 0 (Zero Fact Thm)

cos(x) = −1

2OR cos(x) =

1

2(algebra)

Solve

cos (x) = −1

2

Solution:

xk ≈ 120◦ + k360◦ for k ∈ Z

ORxk ≈ 240◦ + k360◦ for k ∈ Z

Solve

cos (x) =1

2

Solution:

xk ≈ 60◦ + k360◦ for k ∈ Z

ORxk ≈ 300◦ + k360◦ for k ∈ Z

8. Solve6cos2(x) + − 1cos(x) + − 1 = 0

Solution:

6cos2(x) + − 1cos(x) + − 1 = 0 (given)[

3cos(x) + 1]

·[

2cos(x) + − 1]

= 0 (factor)

3cos(x) + 1 = 0 OR 2cos(x) + − 1 = 0 (Zero Fact Thm)

cos(x) = −1

3OR cos(x) =

1

2(algebra)

Solve

cos (x) = −1

3

Solution:

xk ≈ 109.471◦ + k360◦ for k ∈ Z

ORxk ≈ 250.529◦ + k360◦ for k ∈ Z

Solve

cos (x) =1

2

Solution:

xk ≈ 60◦ + k360◦ for k ∈ Z

ORxk ≈ 300◦ + k360◦ for k ∈ Z

9. Solve12sin2(x) + − 5sin(x) + − 3 = 0

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Solution:

12sin2(x) + − 5sin(x) + − 3 = 0 (given)[

3sin(x) + 1]

·[

4sin(x) + − 3]

= 0 (factor)

3sin(x) + 1 = 0 OR 4sin(x) + − 3 = 0 (Zero Fact Thm)

sin(x) = −1

3OR sin(x) =

3

4(algebra)

Solve

sin (x) = −1

3

Solution:

xk ≈ −19.471◦ + k360◦ for k ∈ Z

ORxk ≈ 199.471◦ + k360◦ for k ∈ Z

Solve

sin (x) =3

4

Solution:

xk ≈ 48.59◦ + k360◦ for k ∈ Z

ORxk ≈ 131.41◦ + k360◦ for k ∈ Z

10. Solve4sin2(x) + 8sin(x) + − 5 = 0

Solution:

4sin2(x) + 8sin(x) + − 5 = 0 (given)[

2sin(x) + 5]

·[

2sin(x) + − 1]

= 0 (factor)

2sin(x) + 5 = 0 OR 2sin(x) + − 1 = 0 (Zero Fact Thm)

sin(x) = −5

2OR sin(x) =

1

2(algebra)

Solve

sin (x) = −5

2

no real solution for x

Solve

sin (x) =1

2

Solution:

xk ≈ 30◦ + k360◦ for k ∈ Z

OR

xk ≈ 150◦ + k360◦ for k ∈ Z

11. Solve10cos2(x) + − 18cos(x) + − 4 = 0

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Solution:

10cos2(x) + − 18cos(x) + − 4 = 0 (given)[

5cos(x) + 1]

·[

2cos(x) + − 4]

= 0 (factor)

5cos(x) + 1 = 0 OR 2cos(x) + − 4 = 0 (Zero Fact Thm)

cos(x) = −1

5OR cos(x) = 2 (algebra)

Solve

cos (x) = −1

5

Solution:

xk ≈ 101.537◦ + k360◦ for k ∈ Z

ORxk ≈ 258.463◦ + k360◦ for k ∈ Z

Solve

cos (x) = 2

no real solution for x

12. Solve60cos2(x) + 2cos(x) + − 6 = 0

Solution:

60cos2(x) + 2cos(x) + − 6 = 0 (given)[

6cos(x) + 2]

·[

10cos(x) + − 3]

= 0 (factor)

6cos(x) + 2 = 0 OR 10cos(x) + − 3 = 0 (Zero Fact Thm)

cos(x) = −1

3OR cos(x) =

3

10(algebra)

Solve

cos (x) = −1

3

Solution:

xk ≈ 109.471◦ + k360◦ for k ∈ Z

ORxk ≈ 250.529◦ + k360◦ for k ∈ Z

Solve

cos (x) =3

10

Solution:

xk ≈ 72.542◦ + k360◦ for k ∈ Z

ORxk ≈ 287.458◦ + k360◦ for k ∈ Z

13. Solve3 + sin(x) = 3cos2(x)

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Solution:

3 + sin(x) = 3cos2(x) (given)

3 + sin(x) = 3(

1 − sin2(x))

(famous Pyth identity)

3 + sin(x) = 3 − 3sin2(x) (famous Pyth identity)

3sin2(x) + 1sin(x) = 0 (given)

sin(x)[

3sin(x) + 1]

= 0 (factor)

sin(x) = 0 OR 3sin(x) + 1 = 0 (Zero Fact Thm)

sin(x) = 0 OR sin(x) = −1

3(algebra)

Solve

sin (x) = 0

Solution:

xk = 0◦ + k180◦ for k ∈ Z

Solve

sin (x) = −1

3

Solution:

xk ≈ −19.471◦ + k360◦ for k ∈ Z

ORxk ≈ 199.471◦ + k360◦ for k ∈ Z

14. Solve3 + − cos(x) = 3sin2(x)

Solution:

3 + − cos(x) = 3sin2(x) (given)

3 + − cos(x) = 3(

1 − cos2(x))

(famous Pyth identity)

3 + − cos(x) = 3 − 3cos2(x) (famous Pyth identity)

3cos2(x) + − 1cos(x) = 0 (given)

cos(x)[

3cos(x) + − 1]

= 0 (factor)

cos(x) = 0 OR 3cos(x) + − 1 = 0 (Zero Fact Thm)

cos(x) = 0 OR cos(x) =1

3(algebra)

Solve

cos (x) = 0

Solution:

xk = 90◦ + k180◦ for k ∈ Z

Solve

cos (x) =1

3

Solution:

xk ≈ 70.529◦ + k360◦ for k ∈ Z

ORxk ≈ 289.471◦ + k360◦ for k ∈ Z

c©2007-2009 MathHands.commathhands pg. 9

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15. Solve2 + sin(x) = 2cos2(x)

Solution:

2 + sin(x) = 2cos2(x) (given)

2 + sin(x) = 2(

1 − sin2(x))

(famous Pyth identity)

2 + sin(x) = 2 − 2sin2(x) (famous Pyth identity)

2sin2(x) + 1sin(x) = 0 (given)

sin(x)[

2sin(x) + 1]

= 0 (factor)

sin(x) = 0 OR 2sin(x) + 1 = 0 (Zero Fact Thm)

sin(x) = 0 OR sin(x) = −1

2(algebra)

Solve

sin (x) = 0

Solution:

xk = 0◦ + k180◦ for k ∈ Z

Solve

sin (x) = −1

2

Solution:

xk ≈ −30◦ + k360◦ for k ∈ Z

ORxk ≈ 210◦ + k360◦ for k ∈ Z

16. Solve2 + − 3sin(x) = 2cos2(x)

Solution:

2 + − 3sin(x) = 2cos2(x) (given)

2 + − 3sin(x) = 2(

1 − sin2(x))

(famous Pyth identity)

2 + − 3sin(x) = 2 − 2sin2(x) (famous Pyth identity)

2sin2(x) + − 3sin(x) = 0 (given)

sin(x)[

2sin(x) + − 3]

= 0 (factor)

sin(x) = 0 OR 2sin(x) + − 3 = 0 (Zero Fact Thm)

sin(x) = 0 OR sin(x) =3

2(algebra)

Solvesin (x) = 0

Solution:

xk = 0◦ + k180◦ for k ∈ Z

Solve

sin (x) =3

2

no real solution for x

c©2007-2009 MathHands.commathhands pg. 10

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17. Solve4 + − 3cos(x) = 4sin2(x)

Solution:

4 + − 3cos(x) = 4sin2(x) (given)

4 + − 3cos(x) = 4(

1 − cos2(x))

(famous Pyth identity)

4 + − 3cos(x) = 4 − 4cos2(x) (famous Pyth identity)

4cos2(x) + − 3cos(x) = 0 (given)

cos(x)[

4cos(x) + − 3]

= 0 (factor)

cos(x) = 0 OR 4cos(x) + − 3 = 0 (Zero Fact Thm)

cos(x) = 0 OR cos(x) =3

4(algebra)

Solve

cos (x) = 0

Solution:

xk = 90◦ + k180◦ for k ∈ Z

Solve

cos (x) =3

4

Solution:

xk ≈ 41.41◦ + k360◦ for k ∈ Z

ORxk ≈ 318.59◦ + k360◦ for k ∈ Z

18. Solve6 + − 5cos(x) = 6sin2(x)

Solution:

6 + − 5cos(x) = 6sin2(x) (given)

6 + − 5cos(x) = 6(

1 − cos2(x))

(famous Pyth identity)

6 + − 5cos(x) = 6 − 6cos2(x) (famous Pyth identity)

6cos2(x) + − 5cos(x) = 0 (given)

cos(x)[

6cos(x) + − 5]

= 0 (factor)

cos(x) = 0 OR 6cos(x) + − 5 = 0 (Zero Fact Thm)

cos(x) = 0 OR cos(x) =5

6(algebra)

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Sec. 03 exercises

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Solve

cos (x) = 0

Solution:

xk = 90◦ + k180◦ for k ∈ Z

Solve

cos (x) =5

6

Solution:

xk ≈ 33.557◦ + k360◦ for k ∈ Z

ORxk ≈ 326.443◦ + k360◦ for k ∈ Z

19. Solvecos(

2x)

= cos(

6x)

Solution: note we will use the famous identity:

cos a − cos b = −2 sin

(

a + b

2

)

sin

(

a − b

2

)

cos(

2x)

= cos(

6x)

(given)

cos(

2x)

− cos(

6x)

= 0 (Bi)

−2 sin

(

2x + 6x

2

)

sin

(

2x − 6x

2

)

= 0 (famous id)

−2 sin(

4x)

· sin(

− 2x)

= 0

sin(

4x)

· sin(

− 2x)

= 0 (divide by -2)

sin(

4x)

= 0 OR sin(

− 2x)

= 0 (ZFT)

Solvesin

(

4x)

= 0

Solution:

4xk = 0◦ + k180◦ for k ∈ Z

xk ≈ 45◦k

Solvesin

(

− 2x)

= 0

Solution:

− 2xk = 0◦ + k180◦ for k ∈ Z

xk ≈ −90◦k

finish solving for x and check solutions..

20. Solvecos(

− 2x)

= cos(

6x)

Solution: note we will use the famous identity:

cos a − cos b = −2 sin

(

a + b

2

)

sin

(

a − b

2

)

c©2007-2009 MathHands.commathhands pg. 12

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cos(

− 2x)

= cos(

6x)

(given)

cos(

− 2x)

− cos(

6x)

= 0 (Bi)

−2 sin

(

− 2x + 6x

2

)

sin

(

− 2x − 6x

2

)

= 0 (famous id)

−2 sin(

2x)

· sin(

− 4x)

= 0

sin(

2x)

· sin(

− 4x)

= 0 (divide by -2)

sin(

2x)

= 0 OR sin(

− 4x)

= 0 (ZFT)

Solvesin

(

2x)

= 0

Solution:

2xk = 0◦ + k180◦ for k ∈ Z

xk ≈ 90◦k

Solvesin

(

− 4x)

= 0

Solution:

− 4xk = 0◦ + k180◦ for k ∈ Z

xk ≈ −45◦k

finish solving for x and check solutions..

21. Solvecos(

− 3x)

= cos(

5x)

Solution: note we will use the famous identity:

cos a − cos b = −2 sin

(

a + b

2

)

sin

(

a − b

2

)

cos(

− 3x)

= cos(

5x)

(given)

cos(

− 3x)

− cos(

5x)

= 0 (Bi)

−2 sin

(

− 3x + 5x

2

)

sin

(

− 3x − 5x

2

)

= 0 (famous id)

−2 sin(

x)

· sin(

− 4x)

= 0

sin(

x)

· sin(

− 4x)

= 0 (divide by -2)

sin(

x)

= 0 OR sin(

− 4x)

= 0 (ZFT)

Solvesin

(

x)

= 0

Solution:

xk = 0◦ + k180◦ for k ∈ Z

xk ≈ 180◦k

Solvesin

(

− 4x)

= 0

Solution:

− 4xk = 0◦ + k180◦ for k ∈ Z

xk ≈ −45◦k

c©2007-2009 MathHands.commathhands pg. 13

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finish solving for x and check solutions..

22. Solvecos(

3x)

= cos(

4x)

Solution: note we will use the famous identity:

cos a − cos b = −2 sin

(

a + b

2

)

sin

(

a − b

2

)

cos(

3x)

= cos(

4x)

(given)

cos(

3x)

− cos(

4x)

= 0 (Bi)

−2 sin

(

3x + 4x

2

)

sin

(

3x − 4x

2

)

= 0 (famous id)

−2 sin

(

7

2x

)

· sin

(

−1

2x

)

= 0

sin

(

7

2x

)

· sin

(

−1

2x

)

= 0 (divide by -2)

sin

(

7

2x

)

= 0 OR sin

(

−1

2x

)

= 0 (ZFT)

Solve

sin

(

7

2x

)

= 0

Solution:

7

2xk = 0◦ + k180◦ for k ∈ Z

xk ≈ 51.43◦k

Solve

sin

(

−1

2x

)

= 0

Solution:

−1

2xk = 0◦ + k180◦ for k ∈ Z

xk ≈ −360◦k

finish solving for x and check solutions..

23. Find all solutionscosx = 2 cos2 x

Solution: do NOT divide.. dangerous.. instead..

cosx = 2 cos2 x (given)

cosx − 2 cos2 x = 0 (algebra)

cosx(1 − 2 cosx) = 0 (algebra, factor)

cosx = 0 1 − 2 cosx = 0 (Zero Factor Theorem)

cosx = 0 cosx = 1/2 (algebra..)

c©2007-2009 MathHands.commathhands pg. 14

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then...

The set of all real solutions to cosx = 0 is of the form...

x = 90◦ + k180◦

said differently...x = . . . ,−90◦, 90◦, 270◦, 450◦, . . .

AND

The set of all real solutions to cosx = .5 is of the form...

x = 60.0◦ + k360◦ or x = 300.0◦ + k360◦

said differently...x = . . . ,−60.0◦, 60.0◦, 300.0◦, 420.0◦, . . .

24. Find all solutionscosx = 1 − sin2 x

Solution: ONE way to look at it..

cosx = 1 − sin2 x (given)

(famous idea change all to cosines...)

cosx = cos2 x (pythagoras ID)

cosx − cos2 x = 0 (algebra)

cosx(1 − cosx) = 0 (algebra, factor)

cosx = 0 1 − cosx = 0 (Zero Factor Theorem)

cosx = 0 cosx = 1 (algebra..)

then...

The set of all real solutions to cosx = 0 is of the form...

x = 90◦ + k180◦

said differently...x = . . . ,−90◦, 90◦, 270◦, 450◦, . . .

AND The set of all real solutions to cosx = 1 is of the form...

x = 0.0◦ + k360◦

said differently...x = . . . ,−360.0◦, 0.0◦, 360.0◦, 720.0◦, . . .

25. Find all solutionstan x = sin x

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Solution: ONE way to look at it..

tan x = sinx (given)

(famous idea move all to one side, try to factor.., change to sines n cosines.)

tan x − sin x = 0 (algebra)

sinx

cosx− sin x = 0 (IDS)

sin x

(

1

cosx− 1

)

= 0 (algebra, factor)

sin x = 01

cosx− 1 = 0 (Zero Factor Theorem)

sin x = 0 cosx = 1 (algebra.., note had to mult by cosx)

then...

The set of all real solutions to sinx = 0 is of the form...

x = 180◦k

said differently...x = . . . ,−180◦, 0◦, 180◦, 360◦, . . .

AND The set of all real solutions to cosx = 1 is of the form...

x = 0.0◦ + k360◦

said differently...x = . . . ,−360.0◦, 0.0◦, 360.0◦, 720.0◦, . . .

HOWEVER, because we mult both by cosx, extraneous solutions may have been introduced so each of thesesolutions should be checked and extraneous solutions need to be discarded.

26. Find all solutions

cos(3x + π) =−1

2

27. Find all solutions

cos(40◦ − 2x) =−1

3

28. Find all solutions4

secx − 1− 1 = secx

29. Find all solutionscsc(2x) = − sin2 +1

30. Find all solutionssin(2x) = cos(2x)

31. Find all solutionssin(4x) = cos(2x)

32. Find all solutionscos(5x) = cos(7x)

c©2007-2009 MathHands.commathhands pg. 16

Trigonometry

Sec. 03 exercises

MathHands.com

Marquez

Solution: ONE way to look at it.. is to move all to one side, set to zero.. then try to to change the differenceto a product.. use famous identities.. then use zero factor theorem..

33. (xtra fun..take your time on this one) Find all solutions

sin(5x) = cos(7x)

c©2007-2009 MathHands.commathhands pg. 17