Solve $\sqrt{3}\tan\theta=2\sin\theta$

Why square the equation? $\cos \theta = \frac{\sqrt 3}{2}$ is easier to solve. Also, include the solutions of $\sin \theta = 0$ before canceling $\sin \theta$.

When you square both sides of a equation you get a new equation that has all the solutions of the first but may admit other solutions. That was in the step $\sqrt{3}=2\cos\theta \Rightarrow 3=4\cos^2 \theta$.

Added: Also you can only divide your initial equation by $\sin \theta\neq 0$. But for $\theta=0$ or $\theta=\pm\pi$, $\sin\theta=\tan\theta =0$, which are solutions of your initial equation.

Added 2: In general the equation $A^n=B^n$ has all the solutions of the equation $A=B$ but may have additional ones. This is a consequence of the algebraic identity $$A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+\cdots +AB^{n-2}+B^{n-1}).$$

Dear PUK, you missed $\theta=0$ because you divided the equation by $\tan\theta$ in the first step which is only possible if $\tan\theta$ is nonzero and finite. That's why $\theta=0$ and $\theta=\pi$ and $\theta=-\pi$ for which $\tan\theta$ vanish must be separately checked and indeed, you will find out that the equation is satisfied because $0=0$.

On the other hand, you added the wrong solutions $\pm 5\pi/6$ because their squared cosine is $3/4$, but the cosine itself has a wrong sign, so your squaring created problems. The squaring was unnecessary, as pointed out by lhf as I was writing this sentence, but if you still want to square it, you have to check all the solutions that they work and you will find out that the $\pm 5\pi/6$ solutions don't.