Calculus of a Single Variable, Chapter 3, 3.3, Section 3.3, Problem 47

You need to evaluate the relative extrema of the function, hence, you need to find the solutions to the equation f'(x) = 0 .
You need to determine the first derivative, using the chain rule, such that:
f'(x) = (sin^2 x + sin x)' => f'(x) = 2sin x*cos x + cos x
You need to solve for x the equation f'(x) = 0:
2sin x*cos x + cos x = 0
Factoring out cos x yields:
cos x*(2sin x + 1) = 0
cos x = 0
You need to remember that the cosine function is 0, in interval (0,2pi), for x = pi/2 and x = 3pi/2 .
2 sin x + 1 = 0 => 2sin x = -1 => sin x = -1/2
You need to remember that sine function is negative on intervals (pi,3pi,2) and (3pi/2,2pi).
x = pi + pi/6 => x = (7pi)/6
x = 2pi - pi/6 => x = (11pi)/6
Hence, evaluating the relative extrema of the function yields ((7pi)/6, f((7pi)/6)) and ((11pi)/6,f((11pi)/6)).