$3\sin^2x=\cos^2x;$ $ 0\leq x\leq 2\pi$ Solve for $x$ 
$3\sin^2x=\cos^2x;$ $0\leq x\leq 2\pi$ Solve for $x$:

I honestly have no idea how to start this. Considering I'm going to get a number, I am clueless. I have learned about $\sin$ and $\cos$ but I do not know how to approach this problem. If anyone can go step-by-step with hints. That would be greatly appreciated.  
EDIT:
$$3\sin^2x=1-\sin^2x$$ $$4\sin^2x=1$$ $$\sin^2x=\frac{1}{4}$$ $$\sqrt{\sin^2x}=\sqrt{\frac{1}{4}}$$ $$\sin x=\pm \left(\frac{1}{2}\right)$$

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.
