Math problem help

Question

Solve the following trigonometric equation for all values of x between 0 and 2π:
2cos²x=1+sinx

Answer 1

Use the Pythagorean identity cos² x = 1 - sin² x to get 2(1 - sin² x) = 1 + sin x, that is, 2 - 2sin² x = 1 + sin x. Add 2sin² x - 2 to both sides to obtain 2sin² x + sin x - 1 = 0. Let t = sin x. Then 2t² + t - 1 = 0, or (2t - 1)(t + 1) = 0. Thus t = 1/2 or t = -1. If t = 1/2, then sin x = 1/2, so x = pi/6 or pi - pi/6 = 5pi/6. If t = -1, then sin x = -1, yielding x = 3pi/2. The solution set is then {pi/6, 5pi/6, 3pi/2}.