Calculus of a Single Variable, Chapter 3, 3.2, Section 3.2, Problem 2

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(pi) = f(3pi).
f(pi) = cot (pi/2) = 0
f(3pi) = cot(3pi/2) = 0
Since all the three conditions are valid, you may apply Rolle's theorem:
f'(c)(b-a) = 0
Replacing 3pi for b and pi for a, yields:
f'(c)(3pi-pi) = 0
You need to evaluate f'(c), using chain rule:
f'(c) = (cot(c/2))' = -1/(2sin^2(c/2))
Replacing the found values in equation f'(c)(3pi-pi) = 0.
-2pi/(2sin^2(c/2)) != 0
Hence, in this case, there is no valid value of c for Rolle's theorem to be applied.