Nice equation exercise asked by Dan Sitaru in many math groups solved by using trigonometric subsitution

Exercise:

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

Solution: Let $x=2\cos \theta \,\,\,\,,\,\,\,\,0\le \theta \le \frac{\pi }{2}$ and we know that $2{{\cos }^{2}}\frac{\theta }{2}=1+\cos \theta $

So \[\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+2\cos \theta }}}}+\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+2\cos \theta }}}}=2\sqrt{2}\cos \theta \]

\[\Rightarrow \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\left( 1+\cos \theta  \right)}}}}+\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2\left( 1+\cos \theta  \right)}}}}=2\sqrt{2}\cos \theta \]

\[\Rightarrow \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{4{{\cos }^{2}}\frac{\theta }{2}}}}}+\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{4{{\cos }^{2}}\frac{\theta }{2}}}}}=2\sqrt{2}\cos \theta \]

$\Rightarrow \sqrt{2+\sqrt{2+\sqrt{2+2\cos \frac{\theta }{2}}}}+\sqrt{2-\sqrt{2+\sqrt{2+2\cos \frac{\theta }{2}}}}=2\sqrt{2}\cos \theta $

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

$\Rightarrow \sqrt{2+\sqrt{2+\sqrt{4{{\cos }^{2}}\frac{\theta }{4}}}}+\sqrt{2-\sqrt{2+\sqrt{4{{\cos }^{2}}\frac{\theta }{4}}}}=2\sqrt{2}\cos \theta $

$\Rightarrow \sqrt{2+\sqrt{2+2\cos \frac{\theta }{4}}}+\sqrt{2-\sqrt{2+2\cos \frac{\theta }{4}}}=2\sqrt{2}\cos \theta $

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

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

 $\Rightarrow \sqrt{4{{\cos }^{2}}\frac{\theta }{16}}+\sqrt{4{{\sin }^{2}}\frac{\theta }{16}}=2\sqrt{2}\cos \theta $ $\Rightarrow \cos \frac{\theta }{16}+\sin \frac{\theta }{16}=\sqrt{2}\cos \theta $

$\Rightarrow \frac{\sqrt{2}}{\sqrt{2}}\cos \frac{\theta }{16}+\frac{\sqrt{2}}{\sqrt{2}}\sin \frac{\theta }{16}=\sqrt{2}\cos \theta \Leftrightarrow \sqrt{2}\left( \frac{1}{\sqrt{2}}\cos \frac{\theta }{16}+\frac{1}{\sqrt{2}}\sin \frac{\theta }{16} \right)=\sqrt{2}\cos \theta $

$\Rightarrow \cos \frac{\pi }{4}\cos \frac{\theta }{16}+\sin \frac{\pi }{4}\sin \frac{\theta }{16}=\cos \theta \Leftrightarrow \cos \left( \frac{\pi }{4}-\frac{\theta }{16} \right)=\cos \theta $ $\Rightarrow \frac{\pi }{4}-\frac{\theta }{16}=\pm \theta +2k\pi \,\,\,,\,\,k\in \mathbb{Z}$

$\Rightarrow -\frac{\theta }{16}-\theta =-\frac{\pi }{4}+2k\pi \Leftrightarrow -\frac{17\theta }{16}=-\frac{\pi }{4}+2k\pi \Rightarrow \theta =\frac{16\pi }{4\times 17}-\frac{2\times 16k\pi }{17}\Rightarrow \theta =\frac{4\pi }{17}-\frac{32k\pi }{17}$

$\Rightarrow -\frac{\theta }{16}+\theta =-\frac{\pi }{4}+2k\pi \Leftrightarrow \frac{15\theta }{16}=-\frac{\pi }{4}+2k\pi \Leftrightarrow \theta =-\frac{4\pi }{15}+\frac{32k\pi }{15}$   (rejected ) ,   $k\in \mathbb{Z}$

So $x=2\cos \theta =2\cos \left( \frac{4\pi }{17} \right)=1.4780178344413182$