Given $\tan x+ \tan 2x=\frac{2}{\sqrt{3}}$, find $\tan x\cot 2x$ I can't solve this problem. I tried to find $\tan x$ directly by solving cubic equations but I failed.
The problem is to find $\tan x\cot 2x$ given that
$$\tan x+ \tan 2x=\frac{2}{\sqrt{3}}, \>\>\>\>\>0<x<\pi/4$$
How am I supposed to solve this problem?

$$\frac{2 \tan x}{1-\tan ^2 x}+\tan x=^*\frac{2}{\sqrt{3}}$$.
$$\tan x=-1.45424, \tan x=0.35178, \tan x =2.25716$$
$L=\tan x \cot 2x$ can be written$^{**}$ as $\frac{1}{2} \left(1-\tan ^2 x\right)$ thus possible values are $$L_1=-0.557406,L_2=0.438125,L_3=-2.04739$$

$$(^*)\quad \tan x=u \to u^3-\frac{2 u^2}{\sqrt{3}}-3 u+\frac{2}{\sqrt{3}}=0$$
see solutions here. (press Approximate forms button)
$$(^{**})\quad\tan x \cot 2x=\frac{\sin x\left(\cos ^2 x-\sin ^2 x\right)}{\cos x(2 \sin x \cos x)}=\frac{\cos ^2 x-\sin ^2 x}{2\cos^2 x}=\frac{1}{2}\left(1-\tan^2 x\right)$$
