Answer
$0\leq x\leq \frac{\pi}{6},\frac{5\pi}{6}\leq x\leq 2\pi$
Work Step by Step
$sinx\leq \frac{1}{2}$
For $sinx=\frac{1}{2}$
$.5=sin\frac{\pi}{6}=sin\frac{5 \pi}{6}$
Thus $x=\frac{\pi}{6},\frac{5\pi}{6}$
Hence, the given equation has solutions: $0\leq x\leq \frac{\pi}{6},\frac{5\pi}{6}\leq x\leq 2\pi$
