<s>[INST] Prove that $4\sin\frac{2\pi}{7}- \tan\frac{\pi}{7}= \sqrt{7}$ . 
Prove that $$4\sin\frac{2\pi}{7}- \tan\frac{\pi}{7}= \sqrt{7}$$

I think midway calculations are not easy formulas, I can't find what kind of formula transformation to solve! I think the best solution here is using right triangle . . . I have one solution too, but not pretty . !
 [/INST] Hint:
Rewrite this as
$\tan\frac{\pi}{7}(8\cos^2\frac\pi7 -1)=\sqrt7$
$\tan^2\frac{\pi}{7}(8\cos^2\frac\pi7 -1)^2=7$
$(1-\cos^2\frac{\pi}{7})(8\cos^2\frac\pi7 -1)^2=7\cos^2\frac\pi7$
$(1-\cos\frac{2\pi}{7})(4\cos\frac{2\pi}{7}+3)^2=7\cos\frac{2\pi}{7}+7$
$8\cos^3\frac{2\pi}{7}+4\cos^2\frac{2\pi}{7}-4\cos\frac{2\pi}{7}-1=0$
All we need is to prove that $\cos\frac{2\pi}7$ is a root of $8x^3+4x^2-4x-1=0$
Start from $\cos3(\frac{2\pi}{7})=\cos4(\frac{2\pi}{7})$ ...
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