<s>[INST] 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?
 [/INST] Denote $y = \tan x \cot 2x = \frac{1-\tan^2x}2$ and express $\tan x+ \tan 2x=\frac{2}{\sqrt{3}}$ as a system of equations in $x,y$
$$\tan x+\frac{2\tan x}{1-\tan^2x}=\left(1+\frac1y \right)\tan x=\frac{2}{\sqrt{3}}
$$
Then, eliminate $\tan x$ to get
$$\frac3{y^3} -\frac{13}{y}-6=0$$
which is a depressed cubic equation in $\frac1y$, yielding
$$\tan x \cot 2x=y= \left( \frac{2\sqrt{13}}3 \cos \left( \frac13\cos^{-1} \frac{27}{13\sqrt{13}}\right) \right)^{-1}
$$
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