Suppose that $f(0)=f(2\pi)$. Show that there exists an x such that $f(x)=f(x+\pi)$. I am supposed to show that there exists an $x$ in the interval $[0,\pi]$ such that $f(x)=f(x+\pi)$ by considering another function $g:[0,\pi] \to \mathbb{R}$ defined by $g(x)=f(x)-f(x+\pi)$.
Should I use a concrete example by considering a trig function?


You can consider a specific example if it gives you insight, but it is not sufficient: You're asked to prove this for all functions that satisfy the assumptions.
As far as actually making the proof, try considering the intermediate value theorem. Note that $g(0) = f(0) - f(\pi)$ and $g(\pi) = f(\pi) - f(0) = - (f(0) - f(\pi))$. What can you conclude from this?
