40 решить f(x)=2cosx-sin2x [-p/2; p/2] надо найти максимум и минимум(ymax? ymin? )

f ' (x) = -2sin(x) - 2cos(2x) = 0
sin(x) + 1 - 2sin²(x) = 0
2sin²(x) - sin(x) - 1 = 0
D=1+8=3²
(sin(x))₁;₂ = (1 ± 3) / 4
sin(x) = 1   или   sin(x) = -1/2
x = (π/2) + 2πk, k∈Z
x = (-π/6) + 2πn, n∈Z
x = (-5π/6) + 2πk, k∈Z
на отрезке [-π/2; π/2]:
x = π/2
x = -π/6
у(π/2) = 2cos(π/2) - sin(2π/2) = 0-0 = 0 <--- y_min
у(-π/6) = 2cos(-π/6) - sin(-2π/6) = 2cos(π/6) + sin(π/3) = √3+√3/2 = 1.5√3 <--- y_max
график функции для иллюстрации))