Solving for x in a trig equation I had to solve for $x$ in this equation:
$$
\sin(2x)\tan(x) = \sin(2x)
$$
I was rushing, so maybe I didn't think it through, but at the time the most obvious course of action was to divide both sides by $\sin(2x)$, in order to isolate $\tan(x)$ on one side and $\frac{\sin(2x)}{\sin(2x)}$, or $1$, on the other side. So I got $\tan(x) = 45^\circ$, and went on to list the rest of the solutions in the given range $[-2\pi, 2\pi]$.
However, I got a hefty amount of points marked off, and my method detailed above was apparently the main offender. My teacher was quite cryptic about the exact error I made, but I assume I did something illegal when trying to solve for $x$.
Can anyone shed light on what my error was?
 A: As I'm sure you know you can not divide by zero. The function $\sin2x$ can equal zero so you should not divide by it. Instead you should factorize as follows:
$$\sin2x\tan x=\sin2x$$
$$\sin2x\tan x-\sin2x=0$$
$$\sin2x(\tan x-1)=0$$
Hence: $\sin2x=0$ or $\tan x=1$
Then solve those two separately:
$$\sin2x=0\to2x=-4\pi,-3\pi,-2\pi-\pi,0,\pi,2\pi,3\pi,4\pi$$
or
$$\tan x=1\to x=-\frac{7\pi}{4},-\frac{3\pi}{4},\frac{\pi}{4},\frac{5\pi}{4}$$
so:
$$x=-2\pi,-\frac{7\pi}{4},-\frac{3\pi}{2},-\pi,-\frac{3\pi}{4},-\frac{\pi}{2},0,\frac{\pi}{4},\frac{\pi}{2},\pi,\frac{5\pi}{4},\frac{3\pi}{2},2\pi$$
A: \begin{align}
\sin(2x)\tan(x) &= \sin(2x) \\
\color{red}{\sin(2x) = 0} & \text{ or } \tan(x) = 1 \\
x &= -2\pi,-7\pi/4,-3\pi/2,-\pi,-3\pi/4,-\pi/2,0,\pi/4,\pi/2,\pi,5\pi/4,3\pi/2 \text{ or } 2\pi
\end{align}
A: $$\sin 2x\tan x=\sin 2x$$
$$\sin 2x(\tan x-1)=0$$
$$\sin 2x=0\implies 2x=n\pi\ \ \ or\ \ \ x=\frac{n\pi}{2}$$
or $$\tan x=1\implies x=n\pi+\frac{\pi}{4}$$
Where, $n$ is any integer
For given range $[-2\pi, 2\pi]$, one should get 
$$x=-2\pi, -\frac{3\pi}{2},  -\frac{\pi}{2}, 0, \frac{\pi}{2}, \frac{3\pi}{2}, 2\pi $$
or 
$$x=-\frac{7\pi}{4}, -\frac{3\pi}{4}, \frac{\pi}{4}, \frac{5\pi}{4}$$
