<s>[INST] Solving the equation $\tan{x} + \tan \frac{x}{4}=2$ I am trying to solve the equation $\tan{x} + \tan \frac{x}{4}=2$.
1st attempt: I use the following identity $\tan2x=\frac{2\tan x}{1-\tan^{2}x}$. Then get the following equation
$$
u^5 -2u^4-8u^3+12u^2+3u+2=0,
$$
where $u=\tan \frac{x}{4}$. But i can't find any root for this equation.
2nd attempt: $\tan{4x} + \tan x=2\Rightarrow \sin{4x}\cos{x}+\sin{x}\cos{4x}=2\cos{4x}\cos{x}\Rightarrow \sin{5x}=2\cos{4x}\cos{x}$
$\sin{5x}=2\cos{4x}\cos{x}\Rightarrow \frac{1}{2}\sin{5x}=\cos{4x}\cos{x}\Rightarrow \cos({2k\pi\pm\frac{\pi}{3}})\sin{5x}=\cos{4x}\cos{x}$.
I can't continue from here.
 [/INST] you can use $$\tan(x)=\frac{2\tan(x/2)}{1-\tan^2(x/2)}$$ and
$$\tan(x/2)=\frac{2\tan(x/4)}{1-\tan^2(x/4)}$$
putting all things together and factorizing we get
$$\left( {u}^{2}-4\,u+1 \right)  \left( {u}^{3}+2\,{u}^{2}-3\,u-2
 \right) 
=0$$ where $$u=\tan(x/4)$$
</s>