Answer
$\{\frac{\pi}{2}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6} \}$
Work Step by Step
Step 1. Rewrite the equation as
$2sin(x)cos(x)+cos(x)=0$ or $cos(x)(2sin(x)+1)=0$
Step 2. For $cos(x)=0$, we can find $x=\frac{\pi}{2}, \frac{3\pi}{2}$
Step 3. For $sin(x)=-\frac{1}{2}$, we can find $x=\frac{7\pi}{6}, \frac{11\pi}{6}$
Step 4. The solution set is $\{\frac{\pi}{2}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6} \}$
