Answer
$\frac{\pi}{6}$, $\frac{5\pi}{6}$
Work Step by Step
$2\cos x-3 \tan x=0$
$2\cos x-3*\frac{\sin x}{\cos x}=0$
Multiply both sides by $\cos x$:
$2\cos^2 x-3\sin x=0$
$2(1-\sin^2 x)-3\sin x=0$
$2-2\sin^2 x-3\sin x=0$
$2\sin^2 x+3\sin x-2=0$
$(2\sin x-1)(\sin x+2)=0$
If $2\sin x-1=0$, then $2\sin x=1$, and $\sin x=\frac{1}{2}$. The only solutions in $[0, 2\pi)$ are $\frac{\pi}{6}$ and $\frac{5\pi}{6}$.
If $\sin x+2=0$, then $\sin x=-2$. This is not possible because $\sin x$ cannot be less than -1.
