Answer
$\frac{\pi}{2}, \frac{7\pi}{6}$
Work Step by Step
Step 1. Rewrite the equation as $\frac{1}{2}sin\theta-\frac{\sqrt 3}{2}cos\theta=\frac{1}{2}\longrightarrow cos(\frac{\pi}{3})sin\theta-sin(\frac{\pi}{3})cos\theta=\frac{1}{2}\longrightarrow sin(\theta-\frac{\pi}{3})=\frac{1}{2}$
Step 2. Solve the equation above to get $\theta-\frac{\pi}{3}=2k\pi+\frac{\pi}{6}$ and $\theta-\frac{\pi}{3}=2k\pi+\frac{5\pi}{6}$. Thus $\theta=2k\pi+\frac{\pi}{2}$ and $\theta=2k\pi+\frac{7\pi}{6}$
Step 3. Within $[0,2\pi)$, we have $\theta=\frac{\pi}{2}, \frac{7\pi}{6}$
