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

Question

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

At first I believe that finding an answer to that equation it was going to be an easy problem, since this other equation $g(2x) = g(x)\cdot\cos(x)$ could be solved by $g(x) = \text{sinc}(x) = \frac{\sin(x)}{x}$, but I have already tried many different ansatz unsuccessfully for $f(x)$ (real and complex valued functions), so this kind of innocent equation have already beaten me.

Hope you can explained how to solve it, but first of all, please display the solution to the problem (I need it to test solutions to other related problems), this because I am starting to believe that maybe it don't have a solution, or at least a simple one through standard functions.

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Added later:

Reading your answers I understand now is far from be an easy question as I believe at first sight...

For some of you that ask me about additional characteristics for obtain a solution, I am aiming to find a solution $f(x) \in \mathbb{C}$ analytic, since I was thinking about $f(x)$ as a Fourier transform... but reading the answers it looks I will never find its inverse neither...

As @Sangchul Lee explain in the comments, the equation: $$s(2x) = s(x)\cdot\frac{\sin(x)}{x}$$ have been shown here to have a difficult solution: $$s(x) \propto \prod\limits_{p=1}^\infty \frac{\sin(x/2^p)}{x/2^p}$$

Now, by letting $s(x) = f(x)\cdot q(x)$ such as: $$ s(2x) = f(2x)\cdot q(2x) = f(x)\cdot q(x)\cdot\frac{\sin(x)}{x}$$ And following Wolfram-Alpha here if I make the split as: $$ q(2x) = q(x)/x \Rightarrow q(x) \propto \sqrt{x}\,e^{-\frac{\log^2(x)}{\log(4)}}$$ So given there is an existent $q(x) \in \mathbb{R}$:...

1. It is possible to replace $q(2x)/q(x) = 1/x$ in $ f(2x)\frac{q(2x)}{q(x)} = f(x)\sin(x)\cdot\frac{1}{x}$ and simplify it to $f(2x) = f(x)\sin(x)$??

2. Does it means that the main question have at least as solution $f(x) = c\,\frac{\displaystyle{e^{\log^2(x)/\log(4)}}}{\sqrt{x}}\prod\limits_{p=1}^\infty \frac{\sin(x/2^p)}{x/2^p}$??? (with "$c$" an arbitrary constant)

Answer 1

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.