GET OUT PAPER FOR NOTES!!!

Warm-up (3:30 m)

1. Solve for all solutions graphically: sin3x = –cos2x2. Molly found that the solutions to cos x = 1 are

x = 0 + 2kπ AND x = 6.283 + 2kπ, . Is Molly’s solution correct? Why or why not?

k

sin3x = –cos2x

cos x = 1

• x = 0 + 2kπ• x = 6.283 + 2kπ, k

Solving Trigonometric Equations Algebraically

Inverse Trigonometric Functions

• Remember, your calculator must be in RADIAN mode.

• cos x = 0.6– We can use inverse trig functions to solve for x.

Check the solution graphically

k,πk2927.x927.x

)6.0(cosx

6.0xcos1

Why are there two solutions?

k,πk2356.5xπk2927.x

Let’s consider the Unit CircleWhere is x

(cosine) positive?

“All Students Take Calculus”AS

CT

all ratios are positive

sine is positive

tangent is positive

cosine is positive

cosecant is positive

cotangent is positive

secant is positive

How do we find the other solutions algebraically?

For Cosine For Sine

Calculator Solution

– Calculator Solution

Calculator Solution

π – Calculator Solution

cos x = 0.6

k,πk2927.x927.x

)6.0(cosx

6.0xcos1

πk2356.5x

Your Turn:

• Solve for all solutions algebraically:cos x = – 0.3

sin x = –0.75

Your Turn:

• Solve for all solutions algebraically:sin x = 0.5

What about tangent?

• The solution that you get in the calculator is the only one!

tan x = –5

Your Turn:

• Solve for all solutions algebraically:1. cos x = –0.2 2. sin x = – ⅓

3. tan x = 3 4. sin x = 4

What’s going on with #4?

• sin x = 4

How would you solve for x if…

3x2 – x = 2

So what if we have…

3 sin2x – sin x = 2

What about…

tan x cos2x – tan x = 0

Your Turn:• Solve for all solutions algebraically:5. 4 sin2x = 5 sin x – 1 6. cos x sin2x = cos x

7. sin x tan x = sin x 8. 5 cos2x + 6 cos x = 8

Warm-up (4 m)

1. Solve for all solutions algebraically:3 sin2x + 2 sin x = 5

2. Explain why we would reject the solution cos x = 10

3 sin2x + 2 sin x = 5

Explain why we would reject the solution cos x = 10

What happens if you can’t factor the equation?

• x2 + 5x + 3 = 0

a2ac4bbx

2

The plus or minus symbol means that you

actually have TWO equations!

Quadratic Formula

x2 + 5x + 3 = 0ax2 + bx + c = 0

Using the Quadratic Equation to Solve Trigonometric Equations

• You can’t mix trigonometric functions. (Only one trigonometric function at a time!)

• Must still follow the same basic format:• ax2 + bx + c = 0• 2 cos2x + 6 cos x – 4 = 0• 7 tan2x + 10 = 0

tan2x + 5 tan x + 3 = 0

3 sin2x – 8 sin x = –3

Your Turn:• Solve for all solutions algebraically:

1. sin2x + 2 sin x – 2 = 0

2. tan2x – 2 tan x = 2

3. cos2x = –5 cos x + 1

Seek and Solve!

Remember me?

xtan1xcot

xsin1xcsc

xcos1xsec

xsec1xtan

xcscxcot1

1xcosxsin

22

22

22

Using Reciprocal Identities to Solve Trigonometric Equations

• Our calculators don’t have reciprocal function (sec x, csc x, cot x) keys.

• We can use the reciprocal identities to rewrite

secant, cosecant, and cotangent in terms of cosine, sine, and tangent!

csc x = 2 csc x = ½

cot x cos x = cos x

Your Turn:• Use the reciprocal identities to solve for

solutions algebraically:1. cot x = –10

2. tan x sec x + 3 tan x = 0

3. cos x csc x = 2 cos x

Using Pythagorean Identities to Solve Trigonometric Equations

• You can use a Pythagorean identity to solve a trigonometric equation when:– One of the trig functions is squared– You can’t factor out a GCF– Using a Pythagorean identity helps you rewrite the

squared trig function in terms of the other trig function in the equation

cos2x – sin2x + sin x = 0

sec2x – 2 tan2x = 0

sec2x + tan x = 3

Your Turn:• Use Pythagorean identities to solve for all

solutions algebraically:

1. –10 cos2x – 3 sin x + 9 = 0

2. –6 sin2x + cos x + 5 = 0

3. sec2x + 5 tan x = –2