Answer
$0,\frac{2\pi}{3}, \frac{4\pi}{3}$
Work Step by Step
1. $cos(2\theta)=cos\theta \Longrightarrow 2cos^2\theta-1=cos\theta \Longrightarrow 2cos^2\theta-cos\theta-1=0 \Longrightarrow (2cos\theta+1)(cos\theta-1)=0 \Longrightarrow cos\theta=-\frac{1}{2}, 1$
2. For $cos\theta=1$, we have $\theta=2k\pi$
3. For $cos\theta=-\frac{1}{2}$, we have $\theta=2k\pi+\frac{2\pi}{3}$ or $\theta=2k\pi+\frac{4\pi}{3}$
4. Within $[0,2\pi)$, we have $\theta=0,\frac{2\pi}{3}, \frac{4\pi}{3}$
