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?

\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}
