Verified answer

To identify critical point(s),

0 = f'(x) = -2 sin(x) + 2 cos(2x) => sin(x) = cos(2x) => cos(π/2 - x) = cos(2x), using co-function trig identity sin(u) = cos(π/2 - u); see link below.

Since 0 ≤ x ≤ π/2, π/2 - x = 2x => x = π/6.

f(π/6) = 2 cos(π/6) + sin(π/3) = √3 + (√3)/2 = 3(√3)/2 ≈ 2.6.

Now calculate f(x) at the endpoints of the interval:

f(0) = 2 cos(0) + sin(0) = 2.

f(π/2) = 2 cos(π/2) + sin(π) = 0.

Examining the largest and smallest values of f above gives us the absolute extrema of the function on the interval.

Thus the absolute minimum and maximum values of f on [0, π/2] are 0 and 3(√3)/2, respectively, occurring at x = π/2 and at x = π/6, respectively.