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GET OUT PAPER FOR NOTES!!! Warm-up (3:30 m) 1. Solve for all solutions graphically: sin3x = –cos2x 2. Molly found that the solutions to cos x = 1 are x = 0 + 2kπ AND x = 6.283 + 2kπ, k . Is Molly’s solution correct? Why or why not? 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 cos x 0 . 6 x cos 1 (0 . 6 ) x . 927 x . 927 2 k π , k Why are there two solutions? x . 927 2 k π x 5 . 356 2 k π , k Let’s consider the Unit Circle Where is x (cosine) positive? “All Students Take Calculus” S A sine is positive all ratios are positive cosecant is positive T tangent is positive cotangent is positive C cosine 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 cos x 0 . 6 x cos 1 (0 . 6 ) x . 927 x . 927 2 k π , k x 5 . 356 2 k π 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 Quadratic Formula x b 2 b 4 ac 2a 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: 1. sin2x + 2 sin x – 2 = 0 2. tan2x – 2 tan x = 2 3. cos2x = –5 cos x + 1 Warm-up (4 m) • Solve for all solutions algebraically: 1. tan x cos x + 3 tan x = 0 2. 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. Remember me? sec x 1 cos x csc x 2 1 cot x sin x 2 sin x cos x 1 2 2 1 cot x csc x 2 2 tan x 1 sec x 1 tan x 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 2 cos x – 2 sin x + sin x = 0 2 sec x –2 2 tan x =0 2 sec x + 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