<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] The solutions to $\sin(\frac{x}{2})=0$ or $\cos(\frac{3x}{2})=0$ are given by $$\frac{x}{2} = \pi k \text{ or } \frac{3x}{2} = \frac{\pi}{2}+\pi k, k \in \mathbb Z,$$ which can be rewritten as $$x = 2\pi k \text{ or } x = \frac{\pi}{3}+\frac{2\pi}{3}k, k \in \mathbb Z.$$
The solutions to $\cos(x) = \frac{1}{2}$ or $\sin(x)=0$ are given by $$x = \frac{\pi}{3}+2\pi k, \frac{5\pi}{3}+2\pi k, \pi k, k \in \mathbb Z.$$ These are the same sets of points.
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