1. Aim: Solving Quadratic Trig Equations Course: Alg. 2 & Trig. Aim: How do we solve quadratic trigonometric equations? Do Now: Solve by factoring: x 2 – 3x – 4 = 0

2. Aim: Solving Quadratic Trig Equations Course: Alg. 2 & Trig. Solving Quadratic Equations Solve by factoring: x 2 – 3x – 4 = 0 x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 (x – 4) = 0 (x + 1) = 0 x = 4 x = -1 Solve using quadratic formula: x 2 – 3x – 4 = 0 a = 1, b = -3, c = -4 x = 4 or -1

3. Aim: Solving Quadratic Trig Equations Course: Alg. 2 & Trig. Quadratic Trig Equations - Factoring tan 2  – 3tan  – 4 = 0 (tan  – 4)(tan  + 1) = 0 (tan  – 4) = 0 (tan  + 1) = 0 tan  = 4 tan  = -1 Solve for  by factoring: to nearest degree, in the interval 0º ≤  ≤ 360º tan is (–) in QII & QIV; reference  = 45º tan is (+) in QI & QIII; reference  = 76º QII180 – 45 = 135º QIV360 – 45 = 315º and QI76º QIII180 + 76 = 256º and {76º, 135º, 256º, 315º}  = 45º & 76º

4. Aim: Solving Quadratic Trig Equations Course: Alg. 2 & Trig. Quadratic Trig Equations - Formula tan 2  – 3tan  – 4 = 0 to nearest degree, in the interval 0º ≤  ≤ 360º tan is (–) in QII & QIV; reference  = 45º tan is (+) in QI & QIII; reference  = 76º QII180 – 45 = 135º QIV360 – 45 = 315º and QI76º QIII180 + 76 = 256º and {76º, 135º, 256º, 315º} a = 1, b = -3, c = -4 tan  = 4 and -1  = 45º & 76º

5. Aim: Solving Quadratic Trig Equations Course: Alg. 2 & Trig. Model Problem a = 3, b = -5, c = -4 Given: 3cos 2  – 5cos  – 4 = 0, find  to the nearest degree in the interval 0º ≤  ≤ 360º Calculator 2nd 73 ENTER Display: -.59066722909 ( 5 –))÷ 6 2nd ENTER Display: 53.7956245 (–)ANSCOS 2nd  = 54º

6. Aim: Solving Quadratic Trig Equations Course: Alg. 2 & Trig. Model Problem (Con’t) a = 3, b = -5, c = -4 Given: 3cos 2  – 5cos  – 4 = 0, find  to the nearest degree in the interval 0º ≤  ≤ 360º  = 54º cosine is (–) in QII & QIII; reference  = 54º QII 180 + 54 = 234º and QIII 180 – 54 = 126º {126º, 234º}

7. Aim: Solving Quadratic Trig Equations Course: Alg. 2 & Trig. Special Quadratics Solve for  in the interval 0º ≤  ≤ 360º: tan 2  – 3 = 0  = 60º tan 2  = 3 tan is (–) in QII & QIV; reference  = 60º tan is (+) in QI & QIII; reference  = 60º QII180 – 60 = 120º QIV360 – 60 = 300º and QI60º QIII180 + 60 = 240º and {60º, 120º, 240º, 300º}

8. Aim: Solving Quadratic Trig Equations Course: Alg. 2 & Trig. Model Problem Solve the equation 2cos 2  = cos , for all values of  in the interval 0º ≤  ≤ 360º standard form: 2cos 2  – cos  = 0 cos  (2cos  – 1) = 0 cos  = 0 (2cos  – 1) = 0 factor & solve: 2cos  = 1 cos  = 1/2 or.5  = 90º and 270º  = 60º cosine is (+) in QI & QIV; reference  = 60º QI60º QIV300 – 60 = 300ºand {60º, 90º, 270º, 300º}

9. Aim: Solving Quadratic Trig Equations Course: Alg. 2 & Trig. Model Problem a = 2, b = -1, c = 0 x =.5 and 0 Solve the equation 2cos 2  = cos , for all values of  in the interval 0º ≤  ≤ 360º 2cos 2  – cos  = 0  = 90º and 270º  = 60º cosine is (+) in QI & QIV; reference  = 60º QI60º QIV300 – 60 = 300ºand {60º, 90º, 270º, 300º}

10. Aim: Solving Quadratic Trig Equations Course: Alg. 2 & Trig. Regents Prep Find all values of  in the interval 0 <  < 360 o that satisfy the equation 2 sin 2  + sin  = 1.

11. Aim: Solving Quadratic Trig Equations Course: Alg. 2 & Trig. Model Problem a = 3, b = 4, c = -1 In the interval 0º ≤  ≤ 360º, find to the nearest degree, for all values of  that satisfy the equation rewrite w/o fractions: standard form:  = 12.43º  = 12º

12. Aim: Solving Quadratic Trig Equations Course: Alg. 2 & Trig. Model Problem (Con’t) In the interval 0º ≤  ≤ 360º, find to the nearest degree, for all values of  that satisfy the equation  = 12º sine is (+) in QI & QII; reference  = 12º QI12º QII180 – 12 = 168ºand {12º, 168º}

13. Aim: Solving Quadratic Trig Equations Course: Alg. 2 & Trig.