f:[0,2pi] -> R with f(x)= 0 for x in Q and f(x)=sin(x) for x in R\Q.

At which points c in [0,2pi] is f continuos?

I'm thinking it's discontinuous everywhere except 0,pi and 2pi.

I've come up with this to show discontinuity at the rational points in [0,2pi]

Consider a sequence $\displaystyle x_n \to q \in \mathbb{Q} $ with $\displaystyle x_n \notin \mathbb{Q}$ for every natural n. Then $\displaystyle f(x_n) = sin(x_n) $ does not converge so discontinuous for every $\displaystyle c \in [0,2pi]$ s.t c is rational.

But I'm not sure about how to prove the discontinuity/continuity at other points?