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 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 $
