Is there a function $f(x)$ such that $f(-x) = f(x) + \pi$?

Question

We have: $$e^{-it} = \cos(t) - i \sin(t)$$ $$\cos(t+\pi) = -\cos(t)$$ $$\sin(t+\pi) = -\sin(t)$$

Therefore: $$e^{-it} = -e^{-i(t+\pi)}$$

With a variable $x$, and a function $f(x)$: $$xe^{-if(x)} = (-x)e^{-i(f(x)+\pi)}$$

What $f(x)$ satisfies the above equality? It can be rewritten as: $$f(-x) = f(x) + \pi$$

Thank you.

Answer 1

Suppose $f(-x)=f(x)+\pi$ for all $x$. Then $$ f(x) + f(-x)=f(-x)+\pi+f(x)+\pi $$ so $0=2\pi$. There is no such function.