Answer
$\frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}$
Work Step by Step
Let $sin(x)=\frac{1}{2}$, we can find general forms of the solutions as $x=2k\pi+\frac{\pi}{6}$ or $x=2k\pi+\frac{5\pi}{6}$ where $k$ is an integer. For $k=0,1$, we have $x=\frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}$
