## Trigonometry (11th Edition) Clone

$x = 0, 2\pi, \frac{2\pi}{3}, \frac{4\pi}{3}$
$2\cos^2 x - \cos x = 1$ 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, $x = 0, 2\pi, \frac{2\pi}{3}, \frac{4\pi}{3}$