Answer
$\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{3\pi}{2}$
Work Step by Step
Step 1. Factor the equation as $sin^2(x)(2cos(x)-1)-(2sin(x)-1)=(2sin(x)-1)(sin^2(x)-1)=0$ and use $sin^2(x)-1=-cos^2x$,; we have $(2sin(x)-1)(cos^2(x))=0$, which gives $cos(x)=0$ and $sin(x)=\frac{1}{2}$
Step 2. For $cos(x)=0$, the solutions in $[0,2\pi)$ are $x=\frac{\pi}{2},\frac{3\pi}{2}$
Step 3. For $sin(x)=\frac{1}{2}$, we can find the solutions in $[0,2\pi)$ as $x=\frac{\pi}{6},\frac{5\pi}{6}$
Step 4. The solutions to the original equation are $\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{3\pi}{2}$
