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

$$\begin{align}
\sqrt{7+2\sqrt{7-2\sqrt{7-2x} } } &=x\\
7+2\sqrt{7-2\sqrt{7-2x} }  &=x^2\\
2\sqrt{7-2\sqrt{7-2x} }  &=x^2-7\\
4\left(7-2\sqrt{7-2x} \right)  &=x^4-14x^2+49\\
-8\sqrt{7-2x}   &=x^4-14x^2+21\\
64(7-2x)   &=x^8-28x^6+238x^4-588x^2+441\\
0   &=x^8-28x^6+238x^4-588x^2+128x-7\\
0&=\left(x^2+2x-7\right)\left(x^3-x^2-9x+1\right)^2
\end{align}$$
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.
