Answer
$2k\pi+\frac{2\pi}{3}$, $2k\pi+\frac{4\pi}{3}$
Work Step by Step
Step 1. Given the equation $cos(x)=-\frac{1}{2}$, we have to find the solutions in $[0,2\pi)$ as $x_1=\frac{2\pi}{3}$ and $x_2=\frac{4\pi}{3}$
Step 2. Consider the function has a period of $2\pi$; we can express the solutions as $x=2k\pi+\frac{2\pi}{3}$ and $x=2k\pi+\frac{4\pi}{3}$ where $k$ is an integer.
