
Originally Posted by
red_dog
$\displaystyle \sqrt{21+\sqrt{80}}=\sqrt{(\sqrt{20}+1)^2}=\sqrt{2 0}+1$
$\displaystyle \sqrt{7-\sqrt{21+\sqrt{80}}}=\sqrt{6+\sqrt{20}}=\sqrt{(1+\ sqrt{5})^2}=1+\sqrt{5}$
$\displaystyle \sqrt{7+\sqrt{48}}=\sqrt{(2+\sqrt{3})^2}=2+\sqrt{3 }$
$\displaystyle \sqrt{4-\sqrt{12}}=\sqrt{(\sqrt{3}-1)^2}=\sqrt{3}-1$
Then the equation is
$\displaystyle \sin x=\frac{1+\sqrt{5}}{4}$
We have $\displaystyle \sin\frac{3\pi}{10}=\cos\left(\frac{\pi}{2}-\frac{3\pi}{10}\right)=\cos\frac{2\pi}{10}=1-2\sin^2\frac{\pi}{10}$
$\displaystyle 3\sin\frac{\pi}{10}-4\sin^3\frac{\pi}{10}=1-2\sin^2\frac{\pi}{10}$
Let $\displaystyle \sin\frac{\pi}{10}=t\Rightarrow 4t^3-3t^2-3t+1=0\Rightarrow t_1=1, \ t_2=\frac{\sqrt{5}-1}{4}, \ t_3=\frac{-1-\sqrt{5}}{4}$
But $\displaystyle \sin\frac{\pi}{10}\in(0,1)\Rightarrow \sin\frac{\pi}{10}=\frac{\sqrt{5}-1}{4}$
Then $\displaystyle \sin\frac{3\pi}{10}=1-2\sin^2\frac{\pi}{10}=1-\left(\frac{\sqrt{5}-1}{4}\right)^2=\frac{1+\sqrt{5}}{4}$
Therefore $\displaystyle x=\frac{3\pi}{10}$