Answer
$\frac{\pi}{2},\frac{3\pi}{2}, \frac{\pi}{6},\frac{5\pi}{6}$
Work Step by Step
1. $sin(2\theta)=cos\theta \Longrightarrow 2sin\theta cos\theta=cos\theta \Longrightarrow cos\theta=0,\ or\ sin\theta=\frac{1}{2}$
2. For $cos\theta=0$, we have $\theta=k\pi+\frac{\pi}{2}$
3. For $sin\theta=\frac{1}{2}$, we have $\theta=2k\pi+\frac{\pi}{6}$ or $\theta=2k\pi+\frac{5\pi}{6}$
4. Within $[0,2\pi)$, we have $\theta=\frac{\pi}{2},\frac{3\pi}{2}, \frac{\pi}{6},\frac{5\pi}{6}$
