Aim: How do we solve trig equations involving more than one trig function?

1. Aim: How do we solve trig equations involving more than one trig function? Do Now: Find all values of in the interval 0 <  < 360o that satisfy the equation 2 sin2  + sin  = 1.

2. alternate form of Pythagorean ID cos2 = 1 – sin2  Solving Multiple Trig Function Equations Solve for  in the interval 0º ≤  ≤ 360º: 2cos2  – sin = 1 substitute: 2(1 – sin2 ) – sin  = 1 standard form/ factor/ solve: 2– 2sin2  – sin  = 1 -2sin2  – sin  + 1 = 0 2sin2  + sin  – 1 = 0 (2sin – 1)(sin  + 1) = 0 (2sin – 1) = 0 (sin  + 1) = 0 sin = 1/2 sin  = -1 {30º,150º, 270º}  = 30º or 150º  = 270º

3. alternate form of Pythagorean ID sec2 = 1 + tan2  Solving Multiple Trig Function Equations Solve for  to the nearest degree in the interval 0º ≤  ≤ 360º: 2sec2  – 3tan - 5 = 0 substitute: 2(1 + tan2 ) – 3tan - 5 = 0 standard form/ factor/ solve: 2+ 2tan2  – 3tan - 5 = 0 2tan2  – 3tan - 3 = 0 a = 2, b = -3, c = -3 tan  = 2.19, tan  = -0.69  = 65o, 145o, 245o, 3250

4. Model Problems Find to the nearest degree, all values of  in the interval 00<  < 3600 that satisfy the given equation.

5. Model Problems Find to the nearest degree, all values of  in the interval 00<  < 3600 that satisfy the given equation.