Equation exercise asked by Katrina Linda solved by using by Trigonometric identities and substitution

Exercise:

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

Solution: Let $x=4{{\cos }^{2}}\theta \,\,,\,\,\,\,\,0\le \theta \le \frac{\pi }{2}$  and we know that $2{{\cos }^{2}}\theta =1+\cos 2\theta $

So $4{{\cos }^{2}}\theta +\sqrt{2+\sqrt{2+2\cos \theta }}=2\Leftrightarrow 4{{\cos }^{2}}\theta +\sqrt{2+\sqrt{2\left( 1+\cos \theta  \right)}}=2$

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

$\Leftrightarrow 4{{\cos }^{2}}\theta +2\cos \frac{\theta }{4}=2\Leftrightarrow 4{{\cos }^{2}}\theta =2-2\cos \frac{\theta }{4}\Leftrightarrow 4{{\cos }^{2}}\theta =2\left( 1-\cos \frac{\theta }{4} \right)\Leftrightarrow 4{{\cos }^{2}}\theta =4{{\sin }^{2}}\frac{\theta }{8}$

$\Leftrightarrow {{\sin }^{2}}\left( \frac{\pi }{2}-\theta  \right)={{\sin }^{2}}\left( \frac{\theta }{8} \right)\Leftrightarrow \sin \left( \frac{\pi }{2}-\theta  \right)=\sin \frac{\theta }{8}$

$\Leftrightarrow \frac{\pi }{2}-\theta =\frac{\theta }{8}+2k\pi \Leftrightarrow \frac{\pi }{2}=\frac{\theta }{8}+\theta +2k\pi \Leftrightarrow \frac{\pi }{2}=\frac{9}{8}\theta +2k\pi $

$\Leftrightarrow \theta =\frac{8\pi }{18}+\frac{2\times 8k\pi }{9}\Leftrightarrow \theta =\frac{4\pi }{9}+2k\pi $ thus $x=4{{\cos }^{2}}\left( \frac{4\pi }{9} \right)$ is the only root .