<s>[INST] 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?
 [/INST] How about
$$\tan x-\frac{2\tan x}{1-\tan^2 x}=2\sqrt 3$$
Removing the fraction will give you a cubic equation. At least that is easier than the quartic!
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