Answer
$2k\pi+\frac{7\pi}{6}$, $2k\pi+\frac{11\pi}{6}$
Work Step by Step
Step 1. Given the equation $2sin(x)+1=0$ or $sin(x)=-\frac{1}{2}$, we have to find the solutions in $[0,2\pi)$ as $x_1=\frac{7\pi}{6}$ and $x_2=\frac{11\pi}{6}$
Step 2. Consider the function has a period of $2\pi$; we can express the solutions as $x=2k\pi+\frac{7\pi}{6}$ and $x=2k\pi+\frac{11\pi}{6}$ where $k$ is an integer.
