<s>[INST] What's wrong with this transformation? I have the equation $\tan(x) = 2\sin(x)$ and I'd like to transform it in this way:
$$\tan(x) = 2\sin(x)
\Longleftrightarrow
\frac{\sin(x)}{\cos(x)} = 2\sin(x)
\Longleftrightarrow
\sin(x) = 2\sin(x)\cos(x)
\Longleftrightarrow
\sin(x) = \sin(2x)$$
But I'm getting a wrong result so I suppose that I can't do it in this way. Why?
EDIT: 
This is my solution:
$$\tan(x) = 2\sin(x)
\Longleftrightarrow
\frac{\sin(x)}{\cos(x)} = 2\sin(x)
\Longleftrightarrow
\sin(x) = 2\sin(x)\cos(x)
\Longleftrightarrow
\sin(x) = \sin(2x)
\Longleftrightarrow
\sin(2x) - \sin(x) = 0
\Longleftrightarrow
2\cos(\frac{3x}{2})\sin(\frac{x}{2}) = 0
\Longleftrightarrow
\cos(\frac{3x}{2}) = 0 \vee \sin(\frac{x}{2}) = 0$$
Proper solution is $\cos(x) = \frac{1}{2} \vee \sin(x) = 0$
 [/INST] Equations $\frac{\sin x}{\cos x} = 2\sin x$ and $\sin x =2\sin x\cos x$ are indeed equivalent since for $\cos x = 0$ you have $\sin x =\pm 1$, which doesn't give solution for the second equation.
So, to solve it, we have that either $\sin x = 0$ or $\cos x = \frac 1 2$, thus solutions are given by $x = 2k\pi$, $x = \pi + 2k\pi$, $x = \pm\frac\pi 3 + 2k\pi$, $k\in\mathbb Z$.
To summarize, there is no error in your manipulations.
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