Find the solution of the equation Find all real solutions of this equation :
$$x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$

Using the trick mentioned in a comment, observe that $0\leq x \leq 2$ for this to make sense, so substitute $x = 2\cos(t)$.  Squaring and rearranging, $$4\cos^2(t) - 2 = \sqrt{2-\sqrt{2+2\cos(t)}}.$$  Apply the double angle formula and square again to get $$4\cos^2(2t) - 2 = -\sqrt{2+2\cos(t)}$$ whence $$4\cos(4t)^2-2 = 2\cos(t)$$ and hence $$\cos(8t) = \cos(t).$$
The solutions to this are $8t = \pm t + 2n\pi$, giving $\displaystyle t = \frac{2n\pi}{9}$ or $\displaystyle t = \frac{2n\pi}{7}$.  Restricting $t$ to the range $[0,\pi)$ one finds that only $\displaystyle 2\cos(\frac{2\pi}{9})$ satisfies the original equation.  To check this one need only make sure that the sign of the left hand side matches the appropriate sign on the right at each step.  For instance from the original, $\cos(t)$ must be positive, excluding $\displaystyle t =\frac{8\pi}{9}$ among others, and from the first simplification,  $\cos(2t)$ needs to be positive, which excludes, say, $\displaystyle t = \frac{2\pi}{7}$.  There is exactly one solution since the right hand side of the original is decreasing between $x=0$ and $x=2$ with value $\sqrt{2}$ at $x=2$; this is enough to verify that $\displaystyle 2\cos(\frac{2\pi}{9})$ works after excluding the others.
