<s>[INST] Equation $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$ 
The value of $x$ satisfying the equation $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$ is given by

*

*$2 \cos(10°)$


*$2 \cos(20°)$


*$2 \cos(40°)$


*$2 \cos(80°)$

Using $x=2\cos(\theta)$  we get
$2\cos(\theta)=\sqrt{2+\sqrt{2-\sqrt{2+2\cos\theta}}}$
$2\cos(\theta)=\sqrt{2+\sqrt{2-2\cos\left(\frac{\theta}{2}\right)}}$
$2\cos(\theta)=\sqrt{2+2\sin\left(\frac{\theta}{4}\right)}$
$4\cos^2(\theta)={2+2\sin\left(\frac{\theta}{4}\right)}$
After this step I am not able to solve it
 [/INST] My illustration is outlined below
$\begin{array}{l}
4{\cos ^2}\theta  = 2\left( {1 + \sin \frac{\theta }{4}} \right)\\
4{\cos ^2}{10^o} \ne 2\left( {1 + \sin \frac{{{{10}^o}}}{4}} \right)\\
4{\cos ^2}{10^o} \ne 2\left( {1 + \sin {{2.5}^o}} \right)\\
4{\cos ^2}{10^o} \ne 2\left( {1 + \cos {{87.5}^o}} \right)\\
\\
4{\cos ^2}{20^o} \ne 2\left( {1 + \sin \frac{{{{10}^o}}}{4}} \right)\\
4{\cos ^2}{20^o} \ne 2\left( {1 + \sin {5^o}} \right)\\
4{\cos ^2}{20^o} \ne 2\left( {1 + \cos {{85}^o}} \right)\\
\\
4{\cos ^2}{40^o} = 2\left( {1 + \sin \frac{{{{40}^o}}}{4}} \right)\\
4{\cos ^2}{40^o} = 2\left( {1 + \sin {{10}^o}} \right)\\
4{\cos ^2}{40^o} = 2\left( {1 + \cos {{80}^o}} \right)\\
\\
4{\cos ^2}{80^o} \ne 2\left( {1 + \sin \frac{{{{80}^o}}}{4}} \right)\\
4{\cos ^2}{80^o} \ne 2\left( {1 + \sin {{20}^o}} \right)\\
4{\cos ^2}{80^o} \ne 2\left( {1 + \cos {{70}^o}} \right)
\end{array}$\
Hence the option is $0=40^o$
</s>