Solution to the functional equation $f(2x) = f(x)\cdot\sin(x)$?

Martin has pointed out how one might make wild solutions.

We will show that the only continuous solution to your problem is $f = 0$ when your domain is $\mathbb{R}$.

Indeed note that for all $a \in \mathbb{R}$, $a \not \equiv 0 \mod \pi$

$f(\frac{a}{2}) = \frac{f(a)}{\sin(\frac{a}{2})}$

by induction we have

$f(\frac{a}{2^{n}}) = \frac{f(a)}{\prod_{k=1}^{n}\sin(\frac{a}{2^{k}})}$

Thus if $f(a) \neq 0$ and $f$ is continuous at $0$ we have

$|f(0)| = \lim_{n \rightarrow \infty}|f(\frac{a}{2^{n}})| = \infty$

which is impossible. Thus $f$ is $0$ at all points $a$ such that $a \not \equiv 0 \mod \pi$. By continuity $f$ must be $0$ everywhere.