Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 43

You need to determine the monotony of the function, hence, you need to find where the derivative is positive or negative.
You need to evaluate the first derivative, using the chain rule:
f'(theta) = -2sin theta - 2cos theta*sin theta
You need to solve for theta f'(theta) = 0.
-2sin theta - 2cos theta*sin theta = 0
Factoring out -2 sin theta , yields:
-2sin theta(1 + cos theta) = 0 => -2sin theta = 0 => sin theta = 0 for theta = 0, theta = pi and theta = 2pi
1 + cos theta = 0 => cos theta = -1 for theta = pi
You need to notice that f'(theta)<0 , hence the function decreases, for theta in (0,pi) and f'(theta)>0 , hence, the function increases, for theta in (pi,2pi).
b) The function has a minimum point at theta = pi and it has maximum points at theta = 0 and theta = 2pi.
c) You need to evaluate the inflection points of the function, hence, you need to solve for theta the equation f''(theta) = 0.
f''(theta) =(- 2sin theta - sin 2theta)'
f''(theta) = -2cos theta - 2cos 2theta
f''(theta) = -2cos theta - 2(2cos^2 theta - 1)
f''(theta) = -2cos theta - 4cos^2 theta + 2
You need to solve for theta, f''(theta) = 0 , such that:
-2cos theta - 4cos^2 theta + 2 = 0
2cos^2theta + cos theta - 1 = 0
cos theta = (-1+-sqrt(1+8))/4
cos theta = 1/2 for theta = pi/3 and theta = 5pi/3
or cos theta = -1 for theta = pi
Hence, the function has inflection points at theta = pi/3, theta = pi and theta = 5pi/3.