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. 
 A: Will Jagy's answer has led me to the following approach, which also makes clearer how these cosines come in.
Real solutions $x$ of the given equation satisfy $0\leq x\leq2$. We therefore put $x=2y$ with $0\leq y\leq1$, and obtain the new equation
$$\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-2y}}}}}=2y\ .\tag{1}$$
Introduce the two auxiliary functions
$$c(t):=\sqrt{{1\over2}(1+t)},\qquad s(t):=\sqrt{{1\over2}(1-t)}\ .$$
Equation $(1)$ can then be written as
$$2\ s\circ s\circ c\circ c\circ s(y)=2\ y\ .\tag{2}$$
Let $y=\cos\eta$ with $\eta\in\bigl[0,{\pi\over2}\bigr]$. Then we get in turn
$$\eqalign{
s(y)&=\sin{\eta\over2}=\cos{\pi-\eta\over2},\cr
c\circ s(y)&=\cos{\pi-\eta\over4},\cr
c\circ c\circ s(y)&=\cos{\pi-\eta\over8},\cr
s\circ c\circ c\circ s(y)&=\sin{\pi-\eta\over16}=\cos{7\pi+\eta\over16},\cr
s\circ s\circ c\circ c\circ s(y)&=\sin{7\pi+\eta\over32}=\cos{9\pi-\eta\over32},\cr}$$
whereby all angles appearing on the RHS are in $\bigl[0,{\pi\over2}\bigr]$. 
With $(2)$ we now have
$$\cos{9\pi-\eta\over32}=\cos\eta\ ,$$
and this implies ${9\pi-\eta\over32}=\eta$, or $\eta={3\pi\over11}$. In this way we finally obtain
$$x=2\cos\eta=2\cos{3\pi\over11}=1.30972\ .$$
A: 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
