Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 27

You need to evaluate the points of inflection of the function, hence, you need to solve for x the equation f''(x) = 0.
First, you need to evaluate the first derivative, using chain rule, such that:
f'(x) = sec(x-pi/2)tan(x-pi/2)
You need to evaluate the second derivative, using product rule:
f''(x) = sec'(x-pi/2)tan(x-pi/2) + sec(x-pi/2)tan'(x-pi/2)
f''(x) = sec(x-pi/2)tan^2(x-pi/2) + sec(x-pi/2)(1 +tan^2(x-pi/2))
Factoring out sec(x-pi/2) yields:
f''(x) = sec(x-pi/2)(2tan^2(x-pi/2) + 1)
You need to solve for x equation f''(x) = 0.
sec(x-pi/2) = 0 => 1/(cos(x - pi/2)) = 0 impossible for x in (0,4pi)
(2tan^2(x-pi/2) + 1) = 0 => (2tan^2(x-pi/2) =- 1 impossible for x in (0,4pi)
Hence, evaluating the inflection points in (0,4pi), yields that the function has no inflection points.