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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.

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  • $\begingroup$ It is easy to show that original equation will have 5 roots for $ x\in (0,4\pi)$. $\endgroup$ – samjoe Oct 9 '17 at 14:06
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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 $$u=\tan(x/4)$$

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