Answer
$\{0,\pi \}$
Work Step by Step
Step 1. Factor the equation as $tan^2(x)(cos(x)-1)=0$, which gives two solutions $tan(x)=0$ and $cos(x)=1$
Step 2. For $tan(x)=0$, we can find all x-values in $[0,2\pi)$ as $x=0,\pi$
Step 3. For $cos(x)=1$, we can find all x-values in $[0,2\pi)$ as $x=0$
Step 4. The solutions for the original equation in $[0,2\pi)$ are $\{0,\pi \}$