Calculus of a Single Variable, Chapter 3, 3.2, Section 3.2, Problem 19

The Rolle's theorem is applicable to the given function, only if the function is continuous and differentiable over the interval, and f(a) = f(b). Since all polynomial functions are continuous and differentiable on R, hence, the given function is continuous and differentiable on interval. Now, you need to check if f(0) = f(pi/3).
f(0) = sin(3*0) = 0
f(3) = sin(3*(pi)/3) = sin pi = 0
Since all the three conditions are valid, you may apply Rolle's theorem:
f'(c)(b-a) = 0
Replacing pi/3 for b and 0 for a, yields:
f'(c)(pi/3 - 1) = 0
You need to evaluate f'(c), using chain rule:
f'(c) = (sin(3c))' = (cos 3c)*(3c)' = 3cos 3c
f'(c) = 3cos 3c
Replacing the found values in equation f'(c)(pi/3 - 1) = 0.
(3cos 3c)(pi/3 - 1) = 0 => cos 3c = 0 => 3c = pi/2 or 3c = 3pi/2
c = pi/6 or c = pi/2
Since c = pi/2 does not belong to (0,pi/3), only c = pi/6 is a valid value.
Hence, in this case, the Rolle's theorem may be applied for c = pi/6.