Calculus of a Single Variable, Chapter 3, 3.2, Section 3.2, Problem 22

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(2pi).
f(pi) =sec (pi)= 1/(cos pi) = 1/(-1) = -1
f(2pi) =sec 2pi = 1/(cos 2pi) = 1/1 = 1
Since one of all the three conditions is not valid, f(pi) != f(2pi) , you cannot apply Rolle's theorem.