Calculus of a Single Variable, Chapter 3, 3.2, Section 3.2, Problem 18

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 trigonometric 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(2pi).
f(0) = cos 0 = 1
f(2pi) = cos (2pi) = 1
Since all the three conditions are valid, you may apply Rolle's theorem:
f'(c)(b-a) = 0
Replacing 2pi for b and 0 for a, yields:
f'(c)(2pi-0) = 0
You need to evaluate f'(c):
f'(c) = (cos c)' => f'(c) = - sin c
Replacing the found values in equation f'(c)(2pi-0) = 0
-2pi*sin c = 0 => sin c = 0 => c = 0, c = pi, c = 2pi
Since c = {0,2pi} !in (0,2pi), the only valid value for c is pi .
Hence, in this case, the Rolle's theorem may be applied for c = pi .