<s>[INST] finding the real values of $x$ such that : $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$ How to find the real values of $x$ such that : $$x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$
 [/INST] Here's an approach leading to a "closed formula". 
The two innermost square roots are defined only, if $x\in[-2,2]$, so we can write
$x=2\cos\theta$ for some $\theta\in[0,\pi]$. My solution relies on the trig identities
$$
\sqrt{\frac{1+\cos\alpha}2}=\cos\frac\alpha2
$$
and
$$
\sqrt{\frac{1-\cos\alpha}2}=\sin\frac\alpha2
$$
that hold for all $\alpha\in[0,\pi]$ (outside this range we may get differing signs).
Thus
$$
\sqrt{2+x}=\sqrt{2+2\cos\theta}=2\cos\frac\theta2,
$$
and then
$$
\sqrt{2-\sqrt{2+x}}=\sqrt{2-2\cos\frac\theta2}=2\sin\frac\theta4=2\cos(\frac\pi2-\frac\theta4).
$$
Going on we get
$$
2\cos\theta=x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}=\sqrt{2+2\cos(\frac\pi2-\frac\theta4)}=2\cos(\frac\pi4-\frac\theta8).
$$
In the interval $[0,\pi]$ cosine is injective, so we can conclude that
$$
\theta=\frac\pi4-\frac\theta8\Leftrightarrow\theta=\frac{2\pi}9.
$$
The only real solution is thus
$$
x=2\cos\frac{2\pi}9\approx1.53209.
$$
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