Calculus: Early Transcendentals, Chapter 4, 4.1, Section 4.1, Problem 57

f(t)=2cos(t)+sin(2t)
differentiating,
f'(t)=-2sint+2cos(2t)
Now to find the absolute extrema of the function , that is continuous on a closed interval, we have to find the critical numbers that are in the interval and evaluate the function at the endpoints and at the critical numbers.
Now to find the critical numbers, solve for t for f'(t)=0.
-2sint+2cos(2t)=0
-2sin(t)+2(1-2sin^2(t))=0
-2sint+2-4sin^2(t)=0
-2(2sin^2(t)+sin(t)-1)=0
2sin^2(t)+sin(t)-1=0
sin(t)=(-1+-sqrt(1-(4*2*(-1))))/4
sin(t)=(-1+-3)/4
sin(t)=-1 , 1/2
sin(t)=-1=> t=3pi/2+2*pi*n
sin(t)=1/2=> t=pi/6+2n*pi , (5pi)/6+2n*pi
So , t=pi/6 in the interval (0,pi/2).
Evaluating the function at the critical point pi/6 and at the end points of the interval (o,pi/2),
f(0)=2cos0+sin0=2
f(pi/2)=2cos(pi/2)+sin(pi)=0
f(pi/6)=2cos(pi/6)+sin(pi/3)=(3sqrt(3))/2
So , the function has absolute maximum =(3sqrt(3))/2 , at t=pi/6
It has no absolute minimum, graph is attached.