Answer
$\{\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3} \}$
Work Step by Step
Step 1. Use the identity $cos^2x=1-sin^2x$, we have $3-3sin^2x-sin^2(x)=0$ or $4sin^2x=3$, which gives solutions as $sin(x)=\pm\frac{\sqrt 3}{2}$
Step 2. For $sin(x)=\frac{\sqrt 3}{2}$, we can find all x-values in $[0,2\pi)$ as $x=\frac{\pi}{3},\frac{2\pi}{3}$
Step 3. For $sin(x)=-\frac{\sqrt 3}{2}$, we can find all x-values in $[0,2\pi)$ as $x=\frac{4\pi}{3},\frac{5\pi}{3}$
Step 4. The solutions for the original equation in $[0,2\pi)$ are $\{\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3} \}$
