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

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 $\sqrt{7+2 \sqrt{7}}\approx 3.50592$ !!
