## Calculus: Early Transcendentals 8th Edition

$0\leq x\leq \frac{\pi}{6},\frac{5\pi}{6}\leq x\leq 2\pi$
$sinx\leq \frac{1}{2}$ 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: $0\leq x\leq \frac{\pi}{6},\frac{5\pi}{6}\leq x\leq 2\pi$