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?
 A: 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
$$\tan x \cot 2x=y= \left( \frac{2\sqrt{13}}3 \cos \left( \frac13\cos^{-1} \frac{27}{13\sqrt{13}}\right) \right)^{-1}
$$
A: $$\frac{2 \tan x}{1-\tan ^2 x}+\tan x=^*\frac{2}{\sqrt{3}}$$.
$$\tan x=-1.45424, \tan x=0.35178, \tan x =2.25716$$
$L=\tan x \cot 2x$ 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)
$$(^{**})\quad\tan x \cot 2x=\frac{\sin x\left(\cos ^2 x-\sin ^2 x\right)}{\cos x(2 \sin x \cos x)}=\frac{\cos ^2 x-\sin ^2 x}{2\cos^2 x}=\frac{1}{2}\left(1-\tan^2 x\right)$$
A: If $t:=\tan x$ then $\frac{t(3-t^2)}{1-t^2}=\tfrac{2}{\sqrt{3}}$ so $t^3-\tfrac{2}{\sqrt{3}}t^2+3t+\tfrac{2}{\sqrt{3}}=0$. This has one real root, $t=\tfrac{2+33/a-a}{3\sqrt{3}}$ with $a:=\sqrt[3]{154+9\sqrt{443}}$. Now just calculate $\frac{1-t^2}{2}$.
