Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 42

Given the function f(x)=2sinx+cos(2x) in the interval [0,2pi]
Taking the first derivative we get,
f'(x)=2cosx-2sin(2x)
Inorder to find the critical points we have to equate f'(x)=0.
So we get,
2cosx-2sin(2x)=0
i.e. cosx-2sinxcosx=0
cosx(1-2sinx)=0
implies, cosx=0 and 1-2sinx=0
So cosx=0 implies x=pi/2, 3pi/2 in the interval [0,2pi].
1-2sinx=0 implies, x=pi/6, 5pi/6.
Therefore the critical points are: pi/2, (3pi)/2,pi/6,(5pi)/6
Now taking the second derivative of f(x) we get,
f''(x)=-2sinx-4cos(2x)
Applying second derivative test we have,
f''(pi/2)=4>0
f''(3pi/2)=4>0
f''(pi/6)=-3<0
f''(5pi/6)=-3<0
Therefore the function has relative minima at x=pi/2, (3pi)/2
and the minimum value is f(x)=1
The function has relative maxima at x=pi/6,(5pi)/6
and the maximum value is f(x)=1.5