Characterize entire functions $f$ such that $|f(z)| \leq |\sin(z)|$ [duplicate]

If $|f(z)|\leq|\sin z|$ for all $z$, then the quotient $h(z)=\frac{f(z)}{\sin z}$ is analytic at the complement of the zero set of $\sin z$, as you noted, but it is also analytic at the zeroes (or, rather, has removable singularities there)

Indeed, your inequality implies that whenever $\sin z_0$ is zero, then $f(z_0)$ is zero also: this and a little work will show that $h$ is bounded in a neighborhood of $z_0$.

We thus conclude that $h$ is entire and bounded by $1$. You can probably take it from here.

The function $ h(z) $ must obey $ |h(z)| \le 1 $ on all of $ \mathbb{C} $, which means that $ h(z) $ is a bounded function. By Liouville's theorem, any bounded entire function must be constant, hence $ f(z) = \alpha \sin(z) $ for all $ |\alpha| \le 1 $ characterizes the set of functions you want.