How to solve $\displaystyle x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}$ for $x$? How to solve $\displaystyle x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}$ for $x$?
I tried this way:
Let
$$f(x)=\sqrt{4+\sqrt{4-x}}$$
So,  $x=f^2(x)=f^{2n}(x)$ where $n\in\mathbb{N}$. Then, I tried to prove that $f^k(x)=f^{k+1}(x)$ for any $k\in\mathbb{N}$, but I cannot find any easy way to prove this. If I succeed to prove this, I can write
$$x=\sqrt{4+\sqrt{4-x}}$$
because for $k=0$ I have $f^0(x)=f^1(x)$, but then it will be hard to solve $x=\sqrt{4+\sqrt{4-x}}$.
My question is: how to prove that $f^k(x)=f^{k+1}(x)$ and how to solve $x=\sqrt{4+\sqrt{4-x}}$? If you have any easier method, post your solution.

$\displaystyle x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}$, the replace $x$ at RHS by the same equaility, we get
$$x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}}}}}$$.
Continue in this way then we know that $x$ is equal to the limit(provided it exists) of $$\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\cdots}}}}}}}}$$.
To compute the limit, say $l$, we have $l = \sqrt{4+\sqrt{4-l}}$, which is not hard to solve.
As for existence of the limit, remark that if $a_1(x) = \sqrt{4+\sqrt{4-x}}$ and $a_{n+1}(x) = a_n(a_1(x))$, then $|a_n(x)-a_n(y)| \leq |a_n(4) - a_n(0)|$(since $a_n(x)$ is either increasing or decreasing), which tends to $0$ by repeating using $\sqrt{x}-\sqrt{y} =\frac{x-y}{\sqrt{x}+\sqrt{y}}$. So $a_n(x)$ is Cauchy(since $a_{n+k}(x)= a_n(y)$ for some $y$) and the limit exists
