Find all functions $f: \mathbb{R} \rightarrow \mathbb {R}$ such that $f(x+y)+f(x-y)=2f(x)\cos y$(*) for any x,y real numbers
my attempts: for $x=y=\frac{2\pi}{3}$ in the (*): $f(\frac{2\pi}{3})+f(0)=f(\frac{\pi}{3})$(1) now for $x=0$ and $y=\frac{\pi}{3}$ in the ( *) then $f(\frac{\pi}{3})+f(-\frac{\pi}{3})=f(0)$(2)
(1)+(2) it results that $f(\frac{2\pi}{3})=-f(\frac{\pi}{3})$
now for $x=y=\frac{\pi}{3}$ in the (*) we have $f(\frac{2\pi}{3})+f(0)=f(\frac{\pi}{3})$ so $f(0)=2f(\frac{\pi}{3})$ and also $f(\frac{\pi}{3})=f(-\frac{\pi}{3})$ i dont know if f is a odd function but i have discovered any points with that property $f(x)=f(-x)$ but i dont know how to prove it in general for any x.I don't have any ideas from here.Also i think that the parity of function $cos$ it's important to solve the functional equation.
