## Calculus 8th Edition

$0\leq x\leq\frac{2\pi}{3},\frac{4 \pi}{3}\leq x\leq 2\pi$
$2cosx+1>0$ Here, $2cosx+1=0$ gives $cosx=-\frac{1}{2}$ Since, $cosx<0$ on second and third quadrant thus, x must lie on $\frac{2\pi}{3},\frac{4 \pi}{3}$ Hence, the solution for the given equation is $0\leq x\leq\frac{2\pi}{3},\frac{4 \pi}{3}\leq x\leq 2\pi$