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