Can't solve this trignometric equation, why am I wrong? There is this trig equation:
$$ 
5\tan x - 2\tan 2x = 0 \text{ for 0 < 0 < 360 }
$$
So far I've gotten $$\tan x = \text{0, 180}$$
and all I have to solve now is $$\tan ^2x = 0.2$$ which gives me two angle, $24.1$ and $-24.1$. 
For some reason, this is wrong? Can someone please tell me where I'm going wrong? Thank you in advance!
UPDATE I'm so sorry! I made a mistake in my mathjax, I fixed it.
 A: Answered to original questions:
$$5\tan x - 2\tan^2 x = 0 \iff \tan x(5 -  2\tan x) = 0$$
Than means $5\tan x = 0\iff \tan x = 0\;$ or $\;2\tan x =  5 \iff \tan x = \frac 52$.
Can you take it from here?

A: Answer of the Original Version: 
We have$$\tan x(2\tan x-5)=0$$
If $\tan x=0, x=n180^\circ$ where $n$ is any integer
If $\displaystyle2\tan x-5=0,\tan x=\frac52$
Google says $\displaystyle\arctan\frac52\approx68.1985905^\circ$
$\displaystyle\implies x\approx m180^\circ+68.1985905^\circ$  where $m$ is any integer
Probably you have meant $0<x<360^\circ\implies 0<m180^\circ+68.1985905^\circ<360^\circ$

Answer to the Edited Version:
Using Double Angle formula,
$$5\tan x=2\tan2x=2\frac{2\tan x}{1-\tan^2x}$$
$$\tan x(1-5\tan^2 x)=0$$
If $\tan x=0$ has been dealt already 
$\displaystyle1-5\tan^2x=0\iff \tan^2x=\frac15\implies\cos2x=\frac{1-\tan^2x}{1+\tan^2x}=\frac23$
Using Google, $\displaystyle1\arccos\frac23\approx48.1896851$
$\displaystyle\implies2x=2m\pi\pm48.1896851^\circ\implies x=?$
Find $m$ such that  $0<x<360^\circ$
