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Warm-up (3:30 m) Solve for all solutions graphically: sin3x = –cos2x 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?

sin3x = –cos2x

cos x = 1 x = 0 + 2kπ x = 6.283 + 2kπ,

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

Why are there two solutions? Let’s consider the Unit Circle Where is x (cosine) positive?

“All Students Take Calculus” sine is positive all ratios are positive cosecant is positive T C cosine is positive tangent is positive cotangent is positive secant is positive

How do we find the other solutions algebraically? For Cosine For Sine Calculator Solution – Calculator Solution π – Calculator Solution

cos x = 0.6

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 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? Quadratic Formula x2 + 5x + 3 = 0 The plus or minus symbol means that you actually have TWO equations!

x2 + 5x + 3 = 0 ax2 + 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: sin2x + 2 sin x – 2 = 0 tan2x – 2 tan x = 2 cos2x = –5 cos x + 1

Warm-up (4 m) Solve for all solutions algebraically: tan x cos x + 3 tan x = 0 2 cos2x + 7 cos x – 1 = 0

tan x cos x + 3 tan x = 0

2 cos2x + 7 cos x – 1 = 0

Seek and Solve! You have 30 m to complete the seek and solve. Show all your work on a sheet of paper because I’m collecting it for a classwork grade.

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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: –10 cos2x – 3 sin x + 9 = 0 –6 sin2x + cos x + 5 = 0 sec2x + 5 tan x = –2