$C^{\infty}$ solutions to the functional equation $f(2x)=2\sin(x)f'(x)$

Question

What is a rigorous determination of the $C^\infty$ functions $f:\mathbb{R}\longrightarrow\mathbb{C}$ satisfying $f(2x)=2\sin(x)f'(x)$. Several solutions are informally obtained when plugging in the Fourier series.

Answer 1

I suspect that all solutions are linear combinations of $\sin x$ and $1-\cos x$. That they are solutions to the functional differential equation $$ f(2x)=2f'(x)\sin x \tag{1} $$ can be checked by direct substitution. On the other hand, since Eq. $(1)$ is linear, one can try to solve it using Frobenius method. Writing the solution as a power series, $$ f(x)=x^s\sum_{n=0}^{\infty}a_nx^n, \tag{2} $$ and replacing $\sin x$ with its power series in $(1)$, it's easy to show that the lowest order term of the series $(2)$ must satisfy the equation $$ a_0(2x)^{s}=2sa_0x^s\Rightarrow 2^s = 2s, \tag{3} $$ which has only two real solutions, $s=1$ and $s=2$. This shows that Eq. $(1)$ has at most two linearly independent solutions that can be represented as a power series. The already mentioned functions $\sin x$ and $1-\cos x$ fit the bill.