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