Answer
$\frac{\pi}{2},\frac{7\pi}{6}$
Work Step by Step
1. $sin\theta-\sqrt 3cos\theta=1 \Longrightarrow \frac{1}{2}sin\theta-\frac{\sqrt 3}{2}cos\theta=\frac{1}{2} \Longrightarrow cos\frac{\pi}{3}sin\theta-sin\frac{\pi}{3}cos\theta=\frac{1}{2} \Longrightarrow sin(\theta-\frac{\pi}{3})=\frac{1}{2}$
2. $\theta-\frac{\pi}{3}=2k\pi+\frac{\pi}{6}$ or $\theta-\frac{\pi}{3}=2k\pi+\frac{5\pi}{6}$
3. $\theta=2k\pi+\frac{\pi}{2}$ or $\theta=2k\pi+\frac{7\pi}{6}$
4. Within $[0,2\pi)$, we have $\theta=\frac{\pi}{2},\frac{7\pi}{6}$
