Solve the equation for x if 0 ≤ x < 2π. Give your answer in radians using exact values only. (Enter your answers as a comma-separated list.) cos 2x − 3 sin x − 2 = 0

Answer 1

cos 2x - 3 sin x - 2 = 0

cos (x + x) - 3 sin x - 2 = 0

cos² x - sin² x - 3 sin x - 2 = 0 . . . . now the idea is to express all trig ratios as sin or cos

(1 - sin² x) - sin² x - 3 sin x - 2 = 0

-2 sin² x - 3 sin x - 1 = 0

2 sin² x + 3 sin x + 1 = 0 . . . . . . . . . multiplying through by -1 which gives a quadratic in sin x

(2 sin x + 1) (sin x + 1) = 0

⇒ 2 sin x + 1 = 0 OR sin x + 1 = 0

sin x = -1/2 OR sin x = -1

x = π - π/6 or π + π/6 OR x = π + π/2

Solution set: {5π/6, 7π/6, 3π/2}