continuous and periodic function such that $f(0)=f(2\pi)$.

Question

$f:[0,2\pi]\to\mathbb{R}, $ f continuous. $f(0)=f(2\pi)$ Show that there exists an $x_0\in(0,\pi)$ such that $f(x_0)=f(x_0+\pi)$

Answer 1

Hint: Consider $g:[0, \pi] \to \mathbb{R}$, defined as $g(t) = f(t) - f(t+\pi)$ and use the intermediate value theorem.