Periodic points of $\sin(x)$ on $(-1,1)$

Question

Let $f:(-1,1)\rightarrow (-1,1):$ $$f(x)=\sin(x)$$ Find all the periodic points of $f(x)$.
Note: a point $p$ is called periodic if $\exists n\in\mathbb N: f^n(p)=\underbrace{f\circ f\circ f\circ \dots \circ f}_{ \text{$n$ times }}(p).=p.$

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Obviously, a fixed point of $f$ is a periodic point of $f$ (with period $1$), so
$f(0)=0$ is the only fixed point of $f(x) \implies 0$ is periodic,
I'm not sure how to prove (or disprove) that is unique.
Is it enough to show that $$\big\{ x\in(-1,1): \lim_n f^n(x) = 0 \big\} = (-1,1)?$$ Thank you.

Answer 1

Suppose by contradiction that there exists a non-zero periodic point $x$ of $f$, i.e., $f^p(x)=x$ for some $p\in \mathbb{N}$. Assuming Gary's hint as a fact, then $|\sin^p(x)|=|\sin(\sin^{p-1}(x))|<|\sin^{p-1}(x)|<\ldots <|\sin(x)|<|x|$. Then $|x|<|x|$ contradiction.