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 $\cos(x) = \frac{1}{2} \vee \sin(x) = 0$

There is no problem. Your solution is correct, and the other solution is correct. But you need to do a little work to see that they give the same answers:


*

*$\cos \frac {3x}2 = 0$ gives $x = \frac{(2k+1)\pi}3:\ x \in \{\pm \frac{\pi}3, \pm\pi,\pm \frac{5\pi}3, \pm \frac{7\pi}3, \pm 3\pi, \ldots\}$

*$\sin \frac {x}2 = 0$ gives $x = 2k\pi:\ x \in  \{0, \pm 2\pi, \pm 4\pi, \ldots\}$


while 


*

*$\cos x = \frac 12$ gives $x =\pm \frac{\pi}3 + 2k\pi:\ x \in \{\pm \frac{\pi}3, \pm \frac{5\pi}3, \pm \frac{7\pi}3, \ldots\}$

*$\sin x = 0$ gives $x = k\pi:\ x \in  \{0, \pm\pi, \pm 2\pi, \pm 3\pi,\pm 4\pi, \ldots\}$


Comparison of the values shows that the same ones are in both lists.
