Answer
$0\leq x\leq\frac{2\pi}{3},\frac{4 \pi}{3}\leq x\leq 2\pi$
Work Step by Step
$2cosx+1>0$
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
$0\leq x\leq\frac{2\pi}{3},\frac{4 \pi}{3}\leq x\leq 2\pi$
