Solving $\tan x-\tan(2x)=2\sqrt{3}$ $$\tan x-\tan(2x)=2\sqrt{3}$$

TRY #1
$$\begin{align*}
\tan x-\tan(2x)=2\sqrt{3}&\implies\tan x=2\sqrt{3}+\tan{2x}\\
&\implies \tan^2x=\tan^2(2 x)+4 \sqrt{3} \tan(2 x)+12\\
&\implies\tan^2x=(\frac{2\tan x}{1-\tan^2 x})^2+4\sqrt{3}\frac{2\tan x}{1-\tan^2x}+12
\end{align*}$$
but this will give me an equation with $\tan^4$ which needs quartic formula, too difficult!!

TRY #2
$$\begin{align*}
\tan x-\tan(2x)=2\sqrt{3} &\implies \frac{\sin x}{\cos x}-\frac{\sin 2x}{\cos 2x}=2\sqrt3 \\
&\implies\frac{\sin x\cos 2x-\sin 2x\cos x}{\cos x\cos 2x}=2\sqrt{3}\\
&\implies\frac{-\sin x}{\cos x\cos 2x}=2\sqrt{3}\\
&\implies\frac{-\sin x-2\sqrt{3}\cos x\cos 2x}{1}=0
\end{align*}
$$ then i can't!!
can anyone help me?

let $u = \tan x, \tan 2x = \dfrac{2u}{1-u^2}$  your equation becomes $$u - \dfrac{2u}{1-u^2} = 2\sqrt 3 $$ which can be simplified  $$f(u) = u^3 - 2\sqrt 3 u^2 + u + 2\sqrt 3 = 0 $$ i don't see any simple roots for this. we know that $f(0) = 2\sqrt 3$ and $f(-1) = -2$ so there is at least one $-1 < u < 0, f(u) = 0$
edit:  thanks to user winther, we can factor it. 
$$u^3 - 2\sqrt 3 u^2 + u + 2\sqrt 3 = (u-\sqrt 3)(u^2 -\sqrt 3 u -2) $$ 
