## Precalculus (6th Edition)

$0$, $\frac{2\pi}{3}$, $\frac{4\pi}{3}$
$2\cos^2 x-\cos x=1$ $2\cos^2 x-\cos x-1=0$ $(2\cos x+1)(\cos x-1)=0$ $2\cos x+1=0$ or $\cos x-1=0$ If $2\cos x+1=0$: $2\cos x=-1$ $\cos x=-\frac{1}{2}$ The only solutions in $[0, 2\pi)$ are $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$. If $\cos x-1=0$: $\cos x=1$ The only solution in $[0, 2\pi)$ is $0$.