## Calculus 8th Edition

$0\leq x\leq \frac{\pi}{6},\frac{5\pi}{6}\leq x\leq 2\pi$
$sinx\leq \frac{1}{2}$ For $sinx=\frac{1}{2}$ $sinx=sin\frac{\pi}{6}$ Thus, the complete solutions for $x=\frac{\pi}{6},\frac{5\pi}{6}$ Hence, the given equation has solutions: $0\leq x\leq \frac{\pi}{6},\frac{5\pi}{6}\leq x\leq 2\pi$