Aim: How do we solve quadratic trigonometric equations?

1. Aim: How do we solve quadratic trigonometric equations? Do Now: Solve by factoring: x2 – 3x – 4 = 0

2. Solving Quadratic Equations Solve by factoring: x2 – 3x – 4 = 0 x2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 (x – 4) = 0 (x + 1) = 0 x = 4 x = -1 Solve using quadratic formula: x2 – 3x – 4 = 0 a = 1, b = -3, c = -4 x = 4 or -1

3. Quadratic Trig Equations - Factoring Solve for by factoring: tan2 – 3tan  – 4 = 0 to nearest degree, in the interval 0º ≤  ≤ 360º (tan – 4)(tan + 1) = 0 (tan – 4) = 0 (tan + 1) = 0  = 45º & 76º tan = 4 tan = -1 tan is (–) in QII & QIV; reference  = 45º tan is (+) in QI & QIII; reference  = 76º QII 180 – 45 = 135º QI 76º and and QIV 360 – 45 = 315º QIII 180 + 76 = 256º {76º, 135º, 256º, 315º}

4. Quadratic Trig Equations - Formula tan2 – 3tan  – 4 = 0 to nearest degree, in the interval 0º ≤  ≤ 360º a = 1, b = -3, c = -4 tan = 4 and -1  = 45º & 76º tan is (–) in QII & QIV; reference  = 45º tan is (+) in QI & QIII; reference  = 76º QII 180 – 45 = 135º QI 76º and and QIV 360 – 45 = 315º QIII 180 + 76 = 256º {76º, 135º, 256º, 315º}

5. Calculator ( – 2nd ) ) ÷ ENTER 5 73 6 Display: -.59066722909 2nd COS-1 (–) 2nd ANS ENTER Display: 53.7956245 Model Problem Given: 3cos2 – 5cos  – 4 = 0, find  to the nearest degree in the interval 0º ≤  ≤ 360º a = 3, b = -5, c = -4  = 54º

6. QII QIII Model Problem (Con’t) Given: 3cos2 – 5cos  – 4 = 0, find  to the nearest degree in the interval 0º ≤  ≤ 360º a = 3, b = -5, c = -4  = 54º cosine is (–) in QII & QIII; reference  = 54º 180 – 54 = 126º and {126º, 234º} 180 + 54 = 234º

7. Special Quadratics Solve for  in the interval 0º ≤  ≤ 360º: tan2 – 3 = 0 tan2 = 3  = 60º tan is (–) in QII & QIV; reference  = 60º tan is (+) in QI & QIII; reference  = 60º QII 180 – 60 = 120º QI 60º and and QIV 360 – 60 = 300º QIII 180 + 60 = 240º {60º, 120º, 240º, 300º}

8. Model Problem Solve the equation 2cos2 = cos , for all values of  in the interval 0º ≤  ≤ 360º standard form: 2cos2 – cos = 0 factor & solve: cos(2cos  – 1) = 0 cos = 0 (2cos  – 1) = 0 2cos  = 1 cos  = 1/2 or .5  = 90º and 270º  = 60º cosine is (+) in QI & QIV; reference  = 60º QI 60º and QIV 300 – 60 = 300º {60º, 90º, 270º, 300º}

9. Model Problem Solve the equation 2cos2 = cos , for all values of  in the interval 0º ≤  ≤ 360º 2cos2 – cos = 0 a = 2, b = -1, c = 0 x = .5 and 0  = 90º and 270º  = 60º cosine is (+) in QI & QIV; reference  = 60º QI 60º and QIV 300 – 60 = 300º {60º, 90º, 270º, 300º}

10. Regents Prep Find all values of in the interval 0 <  < 360o that satisfy the equation 2 sin2  + sin  = 1.

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

12. 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 QI 12º and QII 180 – 12 = 168º {12º, 168º}