Equation $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$

Question

The value of $x$ satisfying the equation $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$ is given by

1. $2 \cos(10°)$

2. $2 \cos(20°)$

3. $2 \cos(40°)$

4. $2 \cos(80°)$

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

Answer 1

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 $0=40^o$