Answer
$0,\pi$.
Work Step by Step
Step 1. Rewrite the equation as $sin(x)=\frac{sin(x)}{cos(x)}$ or $sin(x)(cos(x)-1)=0$
Step 2. For $sin(x)=0$, we have solutions in $[0,2\pi)$ as $x=0,\pi$
Step 3. For $cos(x)=1$, we have solutions in $[0,2\pi)$ as $x=0$
Step 4. Thus, we have all the solutions in $[0,2\pi)$ as $x=0,\pi$.
