Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 14

a) You need to determine the intervals where the function increases and decreases, hence, you need to determine where f'(x)>0 or f'(x)<0.
You need to determine the first derivative, using the chain rule:
f'(x) = (cos^2 x - 2sin x)
f'(x) = 2cos x*(-sin x) - 2cos x
You need to solve for x the equation f'(x) = 0:
2cos x*(-sin x) - 2cos x = 0
You need to factor out 2cos x:
2cos x*(-sin x - 1) = 0
2cos x = 0 => cos x = 0 for x = pi/2 and x = 3pi/2
-sin x - 1 = 0 => sin x = - 1 for x = 3pi/2
Since f'(x) >0 for x in (pi/2,3pi/2) , the function increases. Since f'(x)<0 for x in (0,pi/2)U(3pi/2,2pi) yields that the function decreases.
b) The function reaches the extrema at the values of x for f'(x) = 0.
From point a) yields that the function has a maximum point at x = 3pi/2 and a minimum point at x = pi/2 .