Solving $\sin(2x) + 3\cos(2x) = 0$
From $$\sin(2x)\left(1+\frac{3}{\tan(2x)}\right)=0,$$ we conclude that $$\sin(2x)=0\quad OR\quad\tan(2x)=-3.$$ The former is not possible, since then $\tan(2x)=0$, and so $$1+\frac3{\tan(2x)}$$ isn't even defined. Hence, we have $\tan(2x)=-3,$ as you already found through other means.
Before you divide by an expression that may become $0$, you need to separate by cases: one case where that expression is $0$ (which may not be possible), and one where it is not $0$. Let me give you a few examples so you can get the idea.
$$x(x^2+1)=2(x^2+1)$$
Above, we split into the case that $x^2+1=0$ and the case that $x^2+1\neq 0$. In the latter case, division by $x^2+1$ shows us that $x=2$ without difficulty. In the former case, we must solve the equation $x^2+1=0$, which has no real solutions (complex solutions $\pm i$. Thus $x=2$ is the only real solution (and $\pm i$ are the other two complex solutions).
$$2x(x^2-1)=5(x^2-1)$$
Split into the case that $x^2-1=0$ and the case that $x^2-1\neq 0$. In the latter case, division by $x^2-1$ shows us that $x=\frac52$. In the former case, we need to solve $x^2-1=0$ (choose your favorite), and we'll find that $x=\pm1$. Hence, $x=\pm 1$ and $x=\frac52$ are the solutions.
Upshot: Before you divide, make sure that the expression you're dividing by isn't zero! If it can be zero, then deal with that separately.
If $\sin(2x)=0$, then $2x= 0$ or $2x = \pi$. Therefore $\tan(2x) = 0$. And so you're not allowed to divide by $\tan(2x)$.
You can calculate the results with both ways, but you have to check the solutions afterwards.