Find all entire functions such that $|f(z)|\leq |\sin(z)|$ for all $z\in\mathbb C$.

Find all entire functions such that $|f(z)|\leq |\sin(z)|$ for all $z\in\mathbb C$.

Half an hour ago I asked the same question for $|f(z)|\leq |z^2-1|$ here but this time I cannot say something like $$|f(z)|\leq M|z|^m$$ so I don't know how to start. The problem is that $$ \sin(z) = \frac{e^{iz}-e^{-iz}}{2i} $$ gets really large if $z$ gets large.

Hint: Look at the function

$$g(z) = \frac{f(z)}{\sin z}.$$

It is entire and - what?

Bounded. Hence constant. Thus $f(z) = c\cdot\sin z$ for some $c\in\mathbb{C}$ with $\lvert c\rvert \leqslant 1$.

Generally, if you have an inequality $\lvert f(z)\rvert \leqslant C\cdot\lvert g(z)\rvert$ between two meromorphic functions $f$ and $g\not\equiv 0$ on a domain $U\subset\mathbb{C}$, the quotient $\frac{f}{g}$ is a bounded holomorphic function on $U\setminus S$, where $S$ is the union of the pole sets of $f$ and $g$ and the zero set of $g$. As the union of finitely many closed discrete subsets of $U$, it is again a closed discrete subset of $U$, and hence all points of $S$ are removable singularities of $\frac{f}{g}$. It is customary to consider these removed without changing the notation, and so $\frac{f}{g}$ is a bounded holomorphic function on $U$. If $U = \mathbb{C}$ (or if $\mathbb{C}\setminus U$ is "thin enough" - has analytic capacity $0$), it follows that $\frac{f}{g}$ is constant, i.e. $f(z) = c\cdot g(z)$ for some $c$ with $\lvert c\rvert \leqslant C$.