Solutions for the following equation. Let us have an $n$ positive integer. How many solutions do we have for the following equation in the interval $(0,\frac{\pi}{2})$?
$$\underbrace{\cos(\cos(\ldots(\cos x)\ldots))}_{n\text{ times }\cos}=x$$
My idea:
I think that we are looking for a fixed point, and because of Banach-fixed point theorem, there is only one solution, since $x\mapsto cos x$ is a contraction in the interval $(0,\frac{\pi}{2})$. I don't know is this reasoning is enough. And maybe the solution is something else? Thanks for the help!
 A: Let $f^{(1)}(x)=\cos x$ and $f^{(n+1)}(x)=\cos\left(f^{(n)}(x)\right)$. 
For any $x\in I=[0,\pi/2]$ we have:
$$ \lim_{n\to +\infty} f^{(n)}(x)=\xi \tag{1}$$
where $\xi$ is the only root of $x-\cos x$ over $I$, i.e. $\xi=0.739\ldots$.
Quite trivially $\xi$ is also a root of $x-f^{(n)}(x)$ over $I$, and it is the only root, since the existence of another root would imply that the derivative of $x-f^{(n)}(x)$ vanishes for some $x\in I$. That cannot happen since the derivative of $f^{(n)}(x)$ over $I$ is small in absolute value for every $n\geq 2$:
$$\left|\frac{d}{dx}\,f^{(n+1)}(x)\right|=\left|\sin\left(f^{(n)}(x)\right)\cdot\frac{d}{dx}\,f^{(n)}(x)\right|\leq \left|\frac{d}{dx}\,f^{(n)}(x)\right|,$$
$$\left|\frac{d}{dx}\,f^{(2)}(x)\right| = \left|\sin(x)\sin(\cos x)\right|\leq \sin 1=0.84147\ldots.$$
A: In order to apply Banach, you need a $q<1$ such that $|f(x)-f(y)|\le q|x-y|$. For $\cos$ on $(0,\pi)$ we do not have such $q$ (we cannot pick $q$ smaller than $\lim_{x\to\frac\pi2}\cos'(x)=1$). Also, $(0,\frac\pi2)$ is not complete. However,  any $x$ we are looking for is certainly $\in[0,1]$ and here we can let $q=\sin 1<1$ for $\cos$ itself and this also works for the iterated cosine (in fact, $q^n$ will work). But apart fro these minor details your argument is correct: Banach tells us that the iterated cosine has exactly one fixpoint on $[0,1]$. The fixpoint of $\cos$ itself is also a fixedpoitn of the iterated function, and is $\ne0$, hence we have exactly one fixedpoint in $(0,\frac\pi2)$.
