`cos(2x) + sin(x) = 0` Find the exact solutions of the equation in the interval [0, 2pi).

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.`