Find the absolute extrema of 𝑓(𝑥) = 2 sin 𝑥 − cos 2𝑥 on the interval [0, 2pi]

Answer 1

f'(x) = 2cosx + 2sin2x = 0; 2cosx(1 + 2sinx) = 0; cosx = 0; x = π/2 or x = 3π/2

sinx = - 1/2; x = 7π/6 or x = 11π/6

f(0) = - 1

f(π/2) = 3 abs max

f(7π/6) = - 1 - 1/2 = - 3/2 abs min

f(3π/2) = - 1

f(11π/6) = - 3/2 abs min

f(2π) = - 1