Solving $\sin(2x) + 3\cos(2x) = 0$ Okay, there's this simple equation I've been looking into for a while and I don't know why one way of solving it is not correct. See:
$$\sin(2x) + 3\cos(2x) = 0$$
Well, the most obvious would be to rearrange to get:
$$\tan(2x)=-3$$
and get the solutions from there, and this works. The problem comes in when another method is used:
$$\sin(2x) + 3\cos(2x) = 0$$
$$\sin(2x) + \dfrac{3\sin(2x)}{\tan(2x)} = 0$$
$$\sin(2x)\left(1 + \dfrac{3}{\tan(2x)}\right) = 0$$
Resulting in $$\sin(2x) = 0\ or\ \tan(2x)=-3$$
I cannot find anything wrong with this method, except that the solutions that $\sin(2x)$ give are not correct when I tested the results back in the original equation. There's probably something I'm overlooking, but I have to idea what. Could anyone explain? I tried to find an explanation to this on here and on the web but have failed so far.
[Of course, there is also the method of equating $\sin(2x) + 3\cos(2x) = \alpha\sin(x+\mu) = 0$, which confirms the first method, but I really want to know whether there's like an assumption or something I'm ignoring without knowing somewhere in the second method.]
EDIT: Thank you for the answers, but mentioning that dividing by zero is not possible it immediately made me think of something of the like of:
$$2x^2 - 5x = 0$$
If we take $x(2x - 5) = 0$, we get $x = 2.5$ or $x = 0$ which are both correct, but if we go with the 'first method' we get: $$2x^2 = 5x$$ $$2x = 5$$ $$x = 2.5$$ which ignores the existing solution $x = 0$.
reEDIT: Maybe what I'm trying to find is whether there is some way where I can 'see' that using this or that method will not be suitable, so I don't have to go back and reject excess solutions which are not valid?
 A: 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.
A: 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. 
A: Hint:
$$\sin(2x)\left(1 + \dfrac{3}{\tan(2x)}\right) = 0$$
$\sin 2x =0 \implies \dfrac{\sin 2x}{\cos 2x}=0=\tan 2x$, which can't be a solution
of this expression: $$\sin(2x)\left(1 + \dfrac{3}{\tan(2x)}\right)=0$$
A: In the step when writing $\cos(2x)$ as $\cos(2x) \frac{\sin(2x)}{\sin(2x)}=\frac{\sin(2x)}{\tan(2x)}$ you must asume that $\sin(2x)  \neq 0$ so that you don't divide by zero. So in the last step the solution $\sin(2x)=0$ is not involved.
