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 $\cos(x) = \frac{1}{2} \vee \sin(x) = 0$
 A: You must disregard all points where $\cos(x) = 0$.  These are outside of the domain of the initial expression you have on the left-hand side. 
A: 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.
A: 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.
A: To solve, set $\sin(x) = \sin(2x) = 2\sin(x)\cos(x)$.  Subtracting and factoring you get
$$\sin(x)(2\cos(x) - 1) = 0.$$
This happens if $x = 0, \pi, \pi/3, 5\pi/3$.  None of these is a solution to $\cos(x) = 0$, so they all work. 
A: 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.
A: There are two errors in your solution.
First, the transformation $$ 
\frac{\sin x}{\cos x} = 2\sin x \iff \sin x = 2\sin x \cos x
$$ is false.  The correct inference is $$ 
\left( \frac{\sin x}{\cos x} = 2\sin x \text{ and } \cos x \neq 0 \right) \text{ or } \cos x = 0  \iff (\sin x = 2\sin x \cos x \text{ and } \cos x \neq 0) \text{ or } \cos x = 0 
$$ otherwise, going to the left, you divide by zero, and going to the right, you (might) delete the content of your equation by multiplying both sides by zero.  (The hazards are $0 = 0 \implies 0 x = 0 y \not\Rightarrow x = y$ and $x \neq y \not\Rightarrow 0x \neq 0y \implies 0 \neq 0$.)  Starting with either of your equations, you infer the entire other side of the more complicated bi-implication.  Since this can introduce spurious solutions, all solutions must be checked in the original equation at the end of the process.
This suggests a general rule:  Do not cancel.  Instead subtract and factor.  This highlights your second error.  From $$
\sin x = 2 \sin x \cos x \text{,}
$$ subtract and factor to get $$
(1-2\cos x)\sin x = 0 \text{.}
$$  The first factor gives your $\cos x = \frac{1}{2}$.  The second factor gives $\sin x = 0$.  (This is because the product of several things being zero means at least one of them is.)
By proper inference, you should have \begin{align*}
    &      &  \tan x &= 2 \sin x \\
    &\iff  &  \frac{\sin x}{\cos x} &= 2 \sin x \\
    &\iff  &  \sin x &= 2 \sin x \cos x \text{ or } \cos x = 0 \\
    &\iff  &  (1-2\cos x)\sin x &= 0 \text{ or } \cos x = 0 \\ 
    &\iff  &  \cos x &= \frac{1}{2} \text{ or } \sin x = 0 \text{ or } \cos x = 0 \text{.}
\end{align*}  Checking for spurious solutions, all solutions of $\cos x = \frac{1}{2}$ and all solutions of $\sin x = 0$ are solutions of the given equation, but none of the solutions of $\cos x = 0$ are, so the final solution is $\cos x = \frac{1}{2} \text{ or } \sin x = 0$.
Somewhat shorter uses subtract and factor earlier, avoiding the complicated bi-implication altogether: \begin{align*}
    &      &  \tan x &= 2 \sin x \\
    &\iff  &  \frac{\sin x}{\cos x} &= 2 \sin x \\
    &\iff  &  \left( \frac{1}{\cos x} - 2 \right) \sin x &= 0 \\
    &\iff  &  \frac{1}{\cos x} - 2 &= 0 \text{ or } \sin x = 0 \\
    &\iff  &  \frac{1}{\cos x} &= 2 \text{ or } \sin x = 0 \\
    &\iff  &  \cos x &= \frac{1}{2} \text{ or } \sin x = 0 \text{.}
\end{align*} 
