## Calculus: Early Transcendentals 8th Edition

$0\leq x\lt\frac{2\pi}{3},\frac{4 \pi}{3}\lt x\leq 2\pi$
$2cosx+1>0$ Here, $2cosx+1=0$ gives $cosx=-\frac{1}{2}$ cosx is equal to .5 for x=$\frac{2\pi}{3},\frac{4 \pi}{3}$ Hence, the solution for the given equation is $0\leq x\lt\frac{2\pi}{3},\frac{4 \pi}{3}\lt x\leq 2\pi$