solution of nested radical $\sqrt{7+2\sqrt{7-2\sqrt{7-2x} } } =x$ This question is from my friend. he think that there is trigonometry involved to this equation.  
$\sqrt{7+2\sqrt{7-2\sqrt{7-2x} } } =x$
that is from  $\ \ \ x = \sqrt{7+2\sqrt{7-2\sqrt{7-2\sqrt{7+2\sqrt{7-2\sqrt{7-2\sqrt...} } }} } }$  
I try to compare between this nested radical and Ramanujan's nested radicals form ,but  solution is for all positive terms.
Thank you for all comments , thank you so much
 A: $$\begin{align}
\sqrt{7+2\sqrt{7-2\sqrt{7-2x} } } &=x\\
7+2\sqrt{7-2\sqrt{7-2x} }  &=x^2\\
2\sqrt{7-2\sqrt{7-2x} }  &=x^2-7\\
4\left(7-2\sqrt{7-2x} \right)  &=x^4-14x^2+49\\
-8\sqrt{7-2x}   &=x^4-14x^2+21\\
64(7-2x)   &=x^8-28x^6+238x^4-588x^2+441\\
0   &=x^8-28x^6+238x^4-588x^2+128x-7\\
0&=\left(x^2+2x-7\right)\left(x^3-x^2-9x+1\right)^2
\end{align}$$
where a CAS helped with the last factoring. Based on the initial equation, $$\sqrt{7}<x<\sqrt{7+2\sqrt{7}}$$ and you can separately analyze that $x^2+2x-7$ has no zeros in this range. The other factor, $x^3-x^2-9x+1$, has one positive zero in this interval. As the root of a cubic, it is expressible in radicals if you want to use Cardano's method.
If the original equation has a solution at all, it must be this root. It remains to check that this solution actually solves the original equation.
A: I hope and wish that you will get a simpler solution. For the time being, here is mine.
I started writing $$\sqrt{7+2\sqrt{7-2\sqrt{7-2y} } } =x$$ which I solved for $y$; this gives $$y=\frac{1}{128} \left(-x^8+28 x^6-238 x^4+588 x^2+7\right)$$ Replacing $y$ by $x$ and expanding leads to $$x^8-28 x^6+238 x^4-588 x^2+128 x-7=0$$ I suspected that this could be factorized and, after some attempts, a CAS found that this reduces to $$(x^2+2 x-7)(x^3-x^2-9 x+1)^2=0$$ As alex.jordan answered, because of the acceptable range, the solution corresponds to one of the roots of $x^3-x^2-9 x+1=0$. Using Cardano method, we know that there are three real roots. Using the trigonometric method for solving the cubic equation when three real roots are present, we arrive for the only acceptable root to the result $$x=\frac{1}{3}+\frac{4\sqrt{7}}{3}  \cos \left(\frac{1}{3} \tan ^{-1}\left(3
   \sqrt{3}\right)\right)\approx 3.49396$$ which is almost equal to $\frac 72$ and to $\sqrt{7+2 \sqrt{7}}\approx 3.50592$ !!
A: Hint assume it to be $x$ and find the relation between $x$ and $x^2,x^3$ You will get it. If its trigo then it can be related to $\cos\theta$
A: I notice that $ x^3-x^2-9 x+1=0 $ has three roots:
$ x_1=1-4 \cos \left(\frac{\pi }{7}\right)=-2.60388... $
$ x_2=1+4 \cos \left(\frac{2 \pi }{7}\right)=3.49396... $
$ x_3=1-4 \cos \left(\frac{3 \pi }{7}\right)=0.109916... $
Therefore, the above result $ \sqrt{7+2 \sqrt{7-2 \sqrt{7-2 \left(1+4 \cos \left(\frac{2 \pi }{7}\right)\right)}}}=1+4 \cos \left(\frac{2 \pi }{7}\right) $ can be extended to:
$ \sqrt{7-2 \sqrt{7-2 \sqrt{7+2 \left(-1+4 \cos \left(\frac{\pi }{7}\right)\right)}}}=-1+4 \cos \left(\frac{\pi }{7}\right) = -x_1 $
$ \sqrt{7+2 \sqrt{7-2 \sqrt{7-2 \left(1+4 \cos \left(\frac{2 \pi }{7}\right)\right)}}}=1+4 \cos \left(\frac{2 \pi }{7}\right) =x_2 $
$ \sqrt{7-2 \sqrt{7+2 \sqrt{7-2 \left(1-4 \cos \left(\frac{3 \pi }{7}\right)\right)}}}=1-4 \cos \left(\frac{3 \pi }{7}\right) =x_3 $
