Answer
$x = 0, 2\pi, \frac{2\pi}{3}, \frac{4\pi}{3}$
Work Step by Step
$2\cos^2 x - \cos x = 1$
Let $\cos x = x$
$2x^{2} - x = 1$
$2x^{2} - x - 1 = 0$
$2x^{2} - 2x + x - 1 = 0$
$2x(x - 1) + 1(x-1) = 0$
$(2x + 1)(x-1) = 0$
$x = 1, -\frac{1}{2}$
$2x + 1 = 0$
$2x = - 1$
$x = -\frac{1}{2}$
$\cos x = 1$
$x = 0, 2\pi$
$\cos x = -\frac{1}{2}$
$x = \pi - \frac{\pi}{3}, \pi + \frac{\pi}{3}$
$x = \frac{3\pi}{3} - \frac{\pi}{3}, \frac{3\pi}{3} + \frac{\pi}{3}$
$x = \frac{2\pi}{3}, \frac{4\pi}{3}$
Therefore, $x = 0, 2\pi, \frac{2\pi}{3}, \frac{4\pi}{3}$
