A functional equation: $4f(x)^3 +f(3x)=3f(x)$ Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that
  
  $$4f(x)^{3}+f(3x)=3f(x)$$


I know of 2 functions that satisfy the equation but I do not know how to prove that they are the only ones.

Thanks

Andrew
One solution is $f(x)=\sin(x),$ from the identity
$$\sin(3x)=3\sin(x)-4\sin^3(x)\tag{1}$$
which can be checked using the addition formula for sine.

Analytic solutions are (I think) only constants and those of the form $\pm \sin(kx).$ Sketch: Assume $f$ is a nonconstant solution. That $f(0)=0$ follows from the relation, and (this needs work but I think it could be shown) $f$ must be zero for some positive $x.$ Since generally if $f(x)$ is a solution so is $\pm f(kx),$ we may assume $f$ has its first positive zero at $\pi,$ and $f&gt;0$ on $(0,\pi).$ Then the idea is to use the formula $(1)$ to produce the sequence $\pi,\pi/3, \pi/9,...,\pi/3^r,...$ on which $f$ has the same value as $\sin(x).$ We then would have two analytic functions agreeing on a convergent sequence in their domains and could conclude $f(x)=\sin(x)$ for all $x$. The use of the polynomial $g(x)=3x-4x^3$ for this, in getting values of $f(x/3)$ uniquely from values of $f(x),$ is not difficult since $g$ is strictly increasing on $[0,1/2]$ with range $[0,1]$ (one doesn't need $g$ beyond $[0,1/2]$ Of course the first value $f(\pi/3)=\sqrt{3}/2$ comes directly from $f(\pi)=0.$ After that $\pi/9 \approx .349$ is already in the interval $[0,1/2]$ on which $g$ is increasing.
I'm thinking of the Fast Fourier transform. Frequency shift (3x) is easily expressible in FFT domain (1->1 mapping). Multiplication is convolution. Then we get a set of equations. Those convolutions might be quite nasty though.