Here is a very interesting functional equation:
$$f(2x)=4^x f(x)^2f(x+1)^2$$
I found it rather fun to solve, and so I thought I'd share it with the math SE community. Just so you know, I already know the answer to this problem, but I thought the community would find it enjoyable to attempt it.
HINT:
Think about the double-angle formulae of the trigonometric functions.
The following proposes a family of solutions, assuming that $\,f\,$ is continuously differentiable.
Let $\,f(x)=2^{-x+2/3}\,g(x)\,$, then:
$$\require{cancel} \bcancel{2^{-2x+2/3}} g(2x) = \bcancel{2^{2x}}\,\bcancel{2^{-2x+4/3}} g(x)\,\bcancel{2^{-2(x+1)+4/3}}g(x+1) \iff g(2x)=g^2(x)g^2(x+1) $$
Let $\,g(x)=e^{h(x)}\,$, then:
$$ e^{h(2x)} = e^{2h(x)}e^{2h(x+1)} \iff h(2x) = 2 \big(\,h(x)+h(x+1) \,\big) $$
Assuming $\,f\,$ is differentiable then so are $\,g\,$ and $\,h\,$, and:
$$ 2 h'(2x) = 2 \big(\,h'(x)+h'(x+1) \,\big) \iff h'(2x) = h'(x)+h'(x+1) $$
The latter functional equation has the obvious linear solution $\,h'(x)=a(x-1)\,$ with $\,a \in \mathbb{R}\,$.
Tracing the steps back, in the end:
$$ f(x) = 2^{-x+2/3} \cdot c^{x^2 - 2x+2/3} \quad\quad \text{with} \;c \in \mathbb{R}^+ $$