(2у+х)³-(2х-у)³ = 8y³ + 4y²x + 2yx² + x³ - (8x³ - 4x²y + 2xy² - y³) = 8y³ + 4y²x + 2yx² + x³ - 8x³ + 4x²y - 2xy² + y³ = 9y³ + 2y²x + 6yx² - 7x³
Ответ: 9y³ + 2y²x + 6yx² - 7x³
sin(arcCos 0)=sin(П/2)=1.................................
2Sin(2x + π/6) + 1 = √3Sin2x + Cosx
2(Sin2xCosπ/6 + Cos2xSinπ/6) + 1 = √3Sin2x + Cosx
2(Sin2x * √3/2 + Cos2x * 1/2) + 1 = √3Sin2x + Cosx
√3Sin2x + Cos2x + 1 = √3Sin2x + Cosx
Cos2x - Cosx + 1 = 0
2Cos²x - 1 - Cosx + 1 = 0
2Cos²x - Cosx = 0
Cosx(2Cosx - 1) = 0
1) Cosx = 0
x = π/2 + πn , n ∈ Z
2) 2Cosx - 1 = 0
2Cosx = 1
Cosx = 1/2
x= ± arcCos1/2 + 2πn , n ∈ Z
x = ± π/3 + 2πn , n ∈ Z