Hint: $\cos^2 x=1-\sin^2 x$.${}{}{}{}{}{}{}$
Substitute. We get after some simplification $4\sin^2 x=1$. Can you finish from here?
Added: You just need to find $\sin^2 x$, then $\sin x$. You should get $\sin x=\pm\frac{1}{2}$. Then identify the angles from your knowledge about "special angles." One of the angles will turn out to be $\frac{\pi}{6}$, the good old $30^\circ$ angle. There are $3$ others.
Remark: The way you started is fine too, you got $3(1-\cos^2 x)=\cos^2 x$. Now we need to "isolate" $\cos^2 x$. It is easiest to multiply through by $3$ on the left, getting $3-3\cos^2 x=\cos^2 x$. Bring all the $\cos^2 x$ terms to one side. We get $3=4\cos^2 x$, which I prefer to write as $4\cos^2 x=3$. So we get $\cos^2 x=\frac{3}{4}$.
Take the square roots. We get $\cos x=\pm \frac{\sqrt{3}}{2}.$
Following the hint given at the start happens to be a little easier, same principles, nicer numbers.