Precalculus, Chapter 5, 5.5, Section 5.5, Problem 10

You need to evaluate the solution to the equation cos 2x + sin x = 0 , hence, you need to use the formula of double angle for cos 2x , such that:
1 - 2sin^2 x + sin x = 0
You need to re-arrange the terms, such that:
-2sin^2 x + sin x + 1 = 0
2sin^2 x - sin x - 1 = 0
Yo need to replace t for sin x, such that:
2t^2 - t - 1 = 0
Using quadratic formula, yields:
t_1 = (1 + sqrt(1 + 8)/4) => t_1 = (1 + 3)/4 => t_1 = 1
t_2 = (1 - sqrt(1 + 8)/4) => t_2 = (1 - 3)/4 => t_2 = -1/2
You need to replace sin x for t such that:
sin x = 1 => x = pi/2
sin x = -1/2 => x = pi + pi/6 => x = (7pi)/6
sin x = -1/2 => x = 2pi - pi/6 => x = (11pi)/6
Hence, the solution to the equation, in [0,2pi), are x = pi/2, x = (7pi)/6, x = (11pi)/6.