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