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