Find the solution of the equation Find all real solutions of this equation :
$$x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$
 A: 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.
A: WolframAlpha gives a very, very nasty solution.
$$x= \frac{1}{\sqrt[3]{\frac{1}{2}(-1 + i\sqrt{3}})} + \sqrt[3]{\frac{1}{2}(-1 + i\sqrt{3})} \approx 1.5321$$
A: The function 
$$f(x)=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$ 
is non-negative and decreasing on its domain, $-2\le x\le2$.  Since $f(2)=\sqrt2\lt2$, the equation $x=f(x)$ has exactly one solution.  That solution will be a root of the polynomial of degree $8$ that comes from repeated squaring:
$$((x^2-2)^2-2)^2=2+x$$
L.F.'s comment to Nigel Overmars's answer indicates that $x=2\cos(2\pi/9)$ is a solution, so the upshot here is that it is the only solution.
The solution $x=2\cos(2\pi/9)$ can be verified using standard trig identities:
$$\begin{align}
\sqrt{2+2\cos(2\pi/9)}&=2\cos(\pi/9)\\
\sqrt{2-2\cos(\pi/9)}&=2\sin(\pi/18)\\
&=2\cos(\pi/2-\pi/18)\\
&=2\cos(4\pi/9)\\
\sqrt{2+2\cos(4\pi/9)}&=\cos(2\pi/9)
\end{align}$$
A: In order for the equation to hold, we can see that $2-\sqrt{2+x}\geq0$, so $x\leq2$. Clearly, $x\geq0$. Since we are only interested in $\cos t$ and $0\leq\cos t\leq 1$, we may assume $0\leq t\leq\frac\pi2$.
Substitute $x=2\cos t$.
Then: 
$$\begin{align*}&2+x=2(1+\cos t)=4\cos^2\frac t2,\hspace{5pt}\cos\frac t2\geq0 \hspace{5pt}\Rightarrow\hspace{5pt} \sqrt{2+x}=2\cos\frac t2\\
&2-\sqrt{2+x}=2-2\cos\frac t2=2\cdot2\sin^2\frac t4,\hspace{5pt} \sin\frac t4\geq 0 \hspace{5pt}\Rightarrow\hspace{5pt} \sqrt{2-\sqrt{2+x}}=2\sin\frac t4\\
&2+\sqrt{2-\sqrt{2+x}}=2+2\sin\frac t4=2\left(1+\cos\left(\frac\pi2-\frac t4\right)\right)=4\cos^2\left(\frac\pi4-\frac t8\right)
\end{align*}$$
So we get $2\cos t=2\cos\left(\frac\pi4-\frac t8\right)$, i.e. $t=\frac\pi4-\frac t8$ (since if $a=\frac\pi4-\frac t8$ then $0\leq a,t\leq\frac\pi2$, so $\cos a=\cos t$ implies $a=t$). We have $t=\frac 89\left(\frac\pi4\right)=\frac{2\pi}9$. So $x=2\cos\frac{2\pi}9$.
