How to find $x$ in some trigonometric equations How to solve these trigonometric equations?

$$\tan2x-\sin4x = 0$$   

and  

$$\tan2x = \sin x$$

I can't do this, please help me! I did this:  

\begin{align}
\tan2x                &amp;= 2\sin x\\ 
\\
\frac{\sin2x}{\cos2x} &amp;= \tan x
\end{align}
$\sin {2x} = \cfrac {2\tan x} {1+\tan^2 x}$

Just to be clear on how to use this substitution:

$$\tan {2x} = \sin {4x} = \cfrac {2\tan 2x} {1+\tan^2 2x}$$

Then we have $\tan {2x} = 0$ or $1+\tan^2 {2x} = 2$ and in this case $\tan {2x} = \pm  1$. 

... from which point we are reduced to considering simple cases.
Equate tan2x then you will get sinx=sin4x, Now x=n(pi)+((-1)^n)*4x where n is a natural no.
For any sinx=siny we have x=n(pi) + ((-1)^n)*y. you can prove this yourself very easily