Solve the functional equation $f (2x)=f (x)\cos x$ Find all $f: \mathbb R\longrightarrow \mathbb R $ 
such that $f $ is a continuous  function at $0$ and satisfies 
$$\;\forall \:x \in \mathbb R,\; f\left(2x\right) = f\left(x\right)\cos x $$
My try: I just found the $f (x)$ is periodic,  i.e.
$f (2\pi / 2)= f (\pi/2) \cos (\pi /2) $
And$ f (\pi)=f (3\pi)$ ... and so on,
Best I came up with is
$$f (2^n x) = f (x) \cos (x) \cos (2x) ... \cos (2^{n-1} x)$$

We have
$$f(x) = f\left(\dfrac{x}2\right)\cos\left(\dfrac{x}2\right) = f\left(\dfrac{x}4\right)\cos\left(\dfrac{x}4\right)\cos\left(\dfrac{x}2\right)$$
Hence, we have
$$f(x) = f\left(\dfrac{x}{2^{n}}\right) \prod_{k=1}^n \cos\left(\dfrac{x}{2^k}\right) = \dfrac1{2^n} f\left(\dfrac{x}{2^n}\right)\dfrac{\sin(x)}{\sin\left(\dfrac{x}{2^n} \right)}$$
Hence, we have that
$$f(x) = \lim_{n \to \infty} \dfrac1{2^n} f\left(\dfrac{x}{2^n}\right) \dfrac{\sin(x)}{\sin\left(\dfrac{x}{2^n}\right)} = \lim_{n \to \infty}f\left(\dfrac{x}{2^n}\right) \lim_{n \to \infty} \dfrac1{2^n} \dfrac{\sin(x)}{\sin\left(\dfrac{x}{2^n}\right)} = \dfrac{\sin(x)}{x} \lim_{n \to \infty} f\left(\dfrac{x}{2^n}\right)$$
If we assume that $\lim_{x \to 0} f(x)$ exists at the origin, we then have that
$$\lim_{n \to \infty} f\left(\dfrac{x}{2^n}\right) = c$$
from which we obtain that
$$f(x) = c\cdot \dfrac{\sin(x)}x$$
