Дам 40 баллов
помогите решить
f(x)=2cosx-sin2x [-p/2;p/2]
надо найти максимум и минимум(Ymax? Ymin?)

f ' (x) = -2sin(x) - 2cos(2x) = 0sin(x) + 1 - 2sin²(x) = 02sin²(x) - sin(x) - 1 = 0D=1+8=3²(sin(x))₁;₂ = (1 ± 3) / 4sin(x) = 1   или   sin(x) = -1/2x = (π/2) + 2πk, k∈Zx = (-π/6) + 2πn, n∈Zx = (-5π/6) + 2πk, k∈Zна отрезке [-π/2; π/2]:x = π/2x = -π/6у(π/2) = 2cos(π/2) - sin(2π/2) = 0-0 = 0 <--- y_minу(-π/6) = 2cos(-π/6) - sin(-2π/6) = 2cos(π/6) + sin(π/3) = √3+√3/2 = 1.5√3 <--- y_maxграфик функции для иллюстрации))