<s>[INST] 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                &= 2\sin x\\ 
\\
\frac{\sin2x}{\cos2x} &= \tan x
\end{align}
 [/INST] $\tan2x=\sin4x$
$=>\frac{\sin2x}{\cos2x}=2\sin2x \cos2x$
$=>\sin2x(2cos^22x-1)=0$
$=>\sin2x\cos4x=0$
$\sin2x=0=>2x=n\pi=>x=\frac{r\pi}{2}$ for some integer r.
$\cos4x=0=>4x=\frac{(2r+1)\pi}{2}=>x=\frac{(2r+1)\pi}{8}$ for some integer r.
Again, $\sin4x=\sin x$,
$4x=n\pi+(-1)^nx$ for some integer n.
If $n$ is even $=2m(say)=>3x=2m\pi=>x=\frac{2m\pi}{3}$
If n is odd$=2m+1(say)=>5x=(2m+1)\pi=>x=\frac{(2m+1)\pi}{5}$
So we need to resolve r, m in integers such that,
$\frac{r\pi}{2} or \frac{(2r+1)\pi}{8}$ equals to $\frac{2m\pi}{3} or \frac{(2m+1)\pi}{5} $
(1)If $\frac{r\pi}{2} = \frac{2m\pi}{3}=>3r=4m=>r|4=>r=4s$ for some integer s.
$=> x= \frac{r\pi}{2} =\frac{4s\pi}{2} =2s\pi  $ .
(2)If $\frac{r\pi}{2} = \frac{(2m+1)\pi}{5}=>5r=(2m+1)2=>5r=4m+2=>5(r-2)=4(m-2)$
$=>4|(r-2)=>r=4s+2$ for some integer s.
$=> x= \frac{r\pi}{2} = \frac{(4s+2)\pi}{2}=(2s+1)\pi$
(3)If $\frac{(2r+1)\pi}{8}=\frac{2m\pi}{3}=>3(2r+1)=16m$  which is impossible as the LHS is odd, the RHS is even for integral r,m. 
So, there will be no solution in this case.
(4)If $\frac{(2r+1)\pi}{8}=\frac{(2m+1)\pi}{5}=>5(2r+1)=8(2m+1)$  which is again unsolvable in integers. 
So, combining all the 4 cases, $x =  t\pi$ where t is any integer(as t is even in case(1), odd in (2)).
</s>