Equation exercise solved by trigonometric substitution leads to golden ratio root

Exercise:

Solve in$\mathbb{R}$ , $\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}}=x$

Solution: Let $x=2\cos \theta \,,\,\,0\le \theta \le \frac{\pi }{2}$ so $\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-2\cos \theta }}}}=2\cos \theta $

\(\Leftrightarrow \sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2\left( 1-\cos \theta  \right)}}}}=2\cos \theta \) but $2{{\sin }^{2}}\frac{\theta }{2}=1-\cos \theta $

$\Leftrightarrow \sqrt{2+\sqrt{2-\sqrt{2+2\sin \frac{\theta }{2}}}}=2\cos \theta \Leftrightarrow \sqrt{2+\sqrt{2-\sqrt{2\left( 1+\sin \frac{\theta }{2} \right)}}}=2\cos \theta $

$\Leftrightarrow \sqrt{2+\sqrt{2-\sqrt{2\left( 1+\cos \left( \frac{\pi }{2}-\frac{\theta }{2} \right) \right)}}}=2\cos \theta \Leftrightarrow \sqrt{2+\sqrt{2-\sqrt{2\left( 1+\cos \left( \frac{\pi -\theta }{2} \right) \right)}}}=2\cos \theta $

$\Leftrightarrow \sqrt{2+\sqrt{2-2\cos \left( \frac{\pi -\theta }{4} \right)}}=2\cos \theta \Leftrightarrow \sqrt{2+\sqrt{2\left( 1-\cos \left( \frac{\pi -\theta }{4} \right) \right)}}=2\cos \theta $

$\Leftrightarrow \sqrt{2+2\sin \left( \frac{\pi -\theta }{8} \right)}=2\cos \theta \Leftrightarrow \sqrt{2\left( 1+\sin \left( \frac{\pi -\theta }{8} \right) \right)}=2\cos \theta $

$\Leftrightarrow \sqrt{2\left( 1+\cos \left( \frac{\pi }{2}-\frac{\pi -\theta }{8} \right) \right)}=2\cos \theta \Leftrightarrow \sqrt{2\left( 1+\cos \left( \frac{3\pi +\theta }{8} \right) \right)}=2\cos \theta $

$\Leftrightarrow 2\cos \left( \frac{3\pi +\theta }{16} \right)=2\cos \theta \Leftrightarrow \cos \left( \frac{3\pi +\theta }{16} \right)=\cos \theta $

So $\frac{3\pi }{16}+\frac{\theta }{16}=\theta +2k\pi \,\,\,or\,\,\,\,\frac{3\pi }{16}+\frac{\theta }{16}=-\theta +2k\pi $

$\Rightarrow \frac{-15}{16}\theta =-\frac{3\pi }{16}+2k\pi \,\,\,\,or\,\,\,\,\,\frac{17}{16}\theta =-\frac{3\pi }{16}+2k\pi $

$\Rightarrow \theta =\frac{3\times 16}{15\times 16}\pi -\frac{2\times 16k\pi }{15}\,\,\,\,or\,\,\,\theta =\frac{-3\times 16}{17\times 16}\pi +\frac{2\times 16k\pi }{17}$

$\Rightarrow \theta =\frac{\pi }{5}-\frac{32k\pi }{15}\,\,\,or\,\,\,\theta =\frac{-3\pi }{17}+\frac{32k\pi }{17}$ thus $x=2\cos \theta =2\cos \left( \frac{\pi }{5} \right)=\frac{1+\sqrt{5}}{2}$