Equicontinuity of set of functions. The problem: Let $x(t) \in C[0,1]$. Then, define $x_n(t)=cos^nx(t)$ for all $n\in \mathbb{N}$. When is the set $(x_n)_{n=1}^{\infty}$ is equicontinuous in $C[0,1]$? Find a criterion in terms of the function $x(t)$. 
My approach: I know that if the derivatives are uniformly bounded, then the set is equicontinuous. However, $x$ is arbitrary so that I cannot say anything about the derivatives. Unfortunately, I do not know any other idea about approaching this question. Can you help me? Thanks for any help. 
Moreover, does something change when we consider $sin$ instead of $cos$?
 A: The sequence $x_n\left(t\right)$ is uniformly bounded by $1$. According to the Arzelà–Ascoli theorem, it will be equicontinuous if and only if it is pre-compact (its closure is compact), that is, any subsequence has a convergent sub-subsequence.
Fortunately, we know the pointwise limit of the sequence. Granted, if $x\left(t_0\right) = \pi+2\pi k$ for some $k\in\mathbb Z$ then there is no limit, but this can easily be overcome by splitting the sequence into even and odd indices. I'll ignore this issue for the sake of simplicity.
Anyway, if $x\left(t\right)\equiv \pi k$ for some $k\in\mathbb Z$ then clearly the sequence is equicontinuous. Otherwise, there will necessarily be values of $t$ for which $\left| \cos\left(x\left(t\right)\right)\right| < 1$ and therefore $\lim_{n\to\infty} x_n\left(t\right) = 0$. This precludes the possibility of $x\left(t\right)$ ever being equal to any $\pi k$, for this would imply the discontinuity of the pointwise limit and therefore of uniform convergence.
In short summary, it would appear that $x\left(t\right)$ should either be always $\pi k$ or never $\pi k$. There are more details to add to the argument of course. Can you fill them in?
