Решение 1. Область определения y = 2cos(x-п/3) D(y) = R 2. Область значения - 1 ≤ 2cos(x-п/3) ≤ 1 - 1/2 ≤ cos(x-п/3) ≤ 1/2 1) cos(x-п/3) ≥ - 1/2 - arccos(-1/2) + 2πk ≤ x - п/3 ≤ arccos(-1/2) + 2πk, k ∈ Z - 2π/3 + 2πk ≤ x - п/3 ≤ 2π/3 + 2πk, k ∈ Z - 2π/3 + π/3 + 2πk ≤ x ≤ 2π/3 + π/3 + 2πk, k ∈ Z - π/3 + 2πk ≤ x ≤ π + 2πk, k ∈ Z 2) cos(x-п/3) ≤ - 1/2 arccos(-1/2) + 2πk ≤ x - п/3 ≤ 2π - arccos(-1/2) + 2πk, k ∈ Z 2π/3 + 2πk ≤ x - п/3 ≤ 2π - 2π/3 + 2πk, k ∈ Z 2π/3 + 2πk ≤ x - п/3 ≤ 4π/3 + 2πk, k ∈ Z 2π/3 + π/3 + 2πk ≤ x ≤ 4π/3 + π/3 + 2πk, k ∈ Z π + 2πk ≤ x ≤ 5π/3 + 2πk, k ∈ Z
Объяснение:
p⁶+36m¹⁰-12m⁵p³=p^6-12p³m^5 +36m^10=
=(p³)² -2*6p³m^5 +(6m^5)² = ( p³-6m^5)²