Answer
$\frac{11\pi}{6}$
Work Step by Step
Step 1. Rewrite the equation as $sin\theta+\sqrt 3cos\theta=1\longrightarrow\frac{1}{2}sin\theta+\frac{\sqrt 3}{2}cos\theta=\frac{1}{2}\longrightarrow cos(\frac{\pi}{3})sin\theta+sin(\frac{\pi}{3})cos\theta=\frac{1}{2}\longrightarrow sin(\theta+\frac{\pi}{3})=\frac{1}{2}$
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 $\theta=2k\pi-\frac{\pi}{6}$ and $\theta=2k\pi+\frac{\pi}{2}$ (discard this one)
Step 3. Within $[0,2\pi)$ and $\cos\theta\ne0$, we have $\theta=\frac{11\pi}{6}$
