<s>[INST] Find a closed form to the solution of $\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-x}}}}}=x$ Hi I try to solve the following nested radical :
$$\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-x}}}}}=x$$
Miraculously the related polynomials is a quintic .More precisely :
$$ x^5 - x^4 - 4 x^3 + 3 x^2 + 3 x - 1=0$$
I know that we can reduce the quintic to a Bring quintic form and use Jacobi theta function .
My question :
Can we hope to see a closed form with radicals ?
Any helps is greatly appreciated 
Thanks a lot for all your contributions. 
 [/INST] One of the solution that can be obtained easily is by substituting $x = 2\cos\theta$
steps as follows
$\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-x}}}} = x$ will become $\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-2\cos\theta}}}} = 2\cos\theta$
Now by applying Half angle cosine formula we can simplify as follows
$\sqrt{2+\sqrt{2+\sqrt{2-2\sin\frac{\theta}{2}}}} = 2\cos\theta$
$\sqrt{2+\sqrt{2+\sqrt{2-2\cos(\frac{\pi}{2} - \frac{\theta}{2})}}} = 2\cos\theta$
$\sqrt{2+\sqrt{2+2\sin(\frac{\pi}{4} - \frac{\theta}{4})}} = 2\cos\theta$
$\sqrt{2+\sqrt{2+2\cos(\frac{\pi}{2} -\frac{\pi}{4} + \frac{\theta}{4})}} = 2\cos\theta$
$\sqrt{2+2\cos(\frac{\pi}{4} -\frac{\pi}{8} + \frac{\theta}{8})} = 2\cos\theta$
$2\cos(\frac{\pi}{8} -\frac{\pi}{16} + \frac{\theta}{16}) = 2\cos\theta$
Now solving for $\theta$ will give rise to $\frac{\pi}{16} + \frac{\theta}{16} = \theta$
$\frac{16\theta}{16}- \frac{\theta}{16} = \frac{\pi}{16} $
$\theta = \frac{\pi}{15}$. Therefore $x$ becomes $2\cos\frac{\pi}{15}$ or $2\cos12^\circ$
As this is multiple of 3, we can calculate value of $2\cos12^\circ$ which we can represent as a finite nested radical as $\frac{1}{2}\times\sqrt{9+\sqrt5+\sqrt{(30-6\sqrt5)}}$
Refer here
</s>