# Thread: Fantastic Trigo-equation

1. ## Fantastic Trigo-equation

Solve :
$\left\{ \begin{array}{l}
\sin x = \frac{{\sqrt {7 - \sqrt {21 + \sqrt {80} } } }}{{1 + \sqrt {7 + \sqrt {48} } - \sqrt {4 - \sqrt {12} } }} \\
0 < x < \frac{\pi }{2} \\
\end{array} \right.$

2. $\sqrt{21+\sqrt{80}}=\sqrt{(\sqrt{20}+1)^2}=\sqrt{2 0}+1$

$\sqrt{7-\sqrt{21+\sqrt{80}}}=\sqrt{6+\sqrt{20}}=\sqrt{(1+\ sqrt{5})^2}=1+\sqrt{5}$

$\sqrt{7+\sqrt{48}}=\sqrt{(2+\sqrt{3})^2}=2+\sqrt{3 }$

$\sqrt{4-\sqrt{12}}=\sqrt{(\sqrt{3}-1)^2}=\sqrt{3}-1$

Then the equation is

$\sin x=\frac{1+\sqrt{5}}{4}$

We have $\sin\frac{3\pi}{10}=\cos\left(\frac{\pi}{2}-\frac{3\pi}{10}\right)=\cos\frac{2\pi}{10}=1-2\sin^2\frac{\pi}{10}$

$3\sin\frac{\pi}{10}-4\sin^3\frac{\pi}{10}=1-2\sin^2\frac{\pi}{10}$

Let $\sin\frac{\pi}{10}=t\Rightarrow 4t^3-3t^2-3t+1=0\Rightarrow t_1=1, \ t_2=\frac{\sqrt{5}-1}{4}, \ t_3=\frac{-1-\sqrt{5}}{4}$

But $\sin\frac{\pi}{10}\in(0,1)\Rightarrow \sin\frac{\pi}{10}=\frac{\sqrt{5}-1}{4}$

Then $\sin\frac{3\pi}{10}=1-2\sin^2\frac{\pi}{10}=1-\left(\frac{\sqrt{5}-1}{4}\right)^2=\frac{1+\sqrt{5}}{4}$

Therefore $x=\frac{3\pi}{10}$

3. Originally Posted by red_dog
$\sqrt{21+\sqrt{80}}=\sqrt{(\sqrt{20}+1)^2}=\sqrt{2 0}+1$

$\sqrt{7-\sqrt{21+\sqrt{80}}}=\sqrt{6+\sqrt{20}}=\sqrt{(1+\ sqrt{5})^2}=1+\sqrt{5}$

$\sqrt{7+\sqrt{48}}=\sqrt{(2+\sqrt{3})^2}=2+\sqrt{3 }$

$\sqrt{4-\sqrt{12}}=\sqrt{(\sqrt{3}-1)^2}=\sqrt{3}-1$

Then the equation is

$\sin x=\frac{1+\sqrt{5}}{4}$

We have $\sin\frac{3\pi}{10}=\cos\left(\frac{\pi}{2}-\frac{3\pi}{10}\right)=\cos\frac{2\pi}{10}=1-2\sin^2\frac{\pi}{10}$

$3\sin\frac{\pi}{10}-4\sin^3\frac{\pi}{10}=1-2\sin^2\frac{\pi}{10}$

Let $\sin\frac{\pi}{10}=t\Rightarrow 4t^3-3t^2-3t+1=0\Rightarrow t_1=1, \ t_2=\frac{\sqrt{5}-1}{4}, \ t_3=\frac{-1-\sqrt{5}}{4}$

But $\sin\frac{\pi}{10}\in(0,1)\Rightarrow \sin\frac{\pi}{10}=\frac{\sqrt{5}-1}{4}$

Then $\sin\frac{3\pi}{10}=1-2\sin^2\frac{\pi}{10}=1-\left(\frac{\sqrt{5}-1}{4}\right)^2=\frac{1+\sqrt{5}}{4}$

Therefore $x=\frac{3\pi}{10}$
hello Thank you
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