![x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz= \\ = (x^2z+2xyz+y^2z)+(x^2y+xy^2)+(xz^2+yz^2)= \\ = z(x^2+2xy+y^2)+xy(x+y)+z^2(x+y)=\\ = z(x+y)^2+xy(x+y)+z^2(x+y)=\\ = z(x+y)(x+y+z)+xy(x+y)=(x+y)(z^2+zx+zy+xy)=\\= (x+y)[z(z+x)+y(z+x)]=(x+y)(z+x)(z+y).](/tpl/images/0409/9446/e1fcc.png)
cos5x + cosx + 2cos2x = 0
2cos(5x+x/2)cos(5x-x/2) + 2cos2x = 0
2cos(6x/2)cos(4x/2) + 2cos2x = 0
2cos3x × cos2x + 2cos2x = 0
2cos2x × (cos3x + 1) = 0 | : 2
cos2x × (cos3x + 1) = 0
cos2x = 0 или cos3x + 1 = 0
2x = π/2 + πn cos3x = -1
x₁ = π/2 × 1/2 + πn × 1/2 3x = π + 2πn
x₁ = π/4 + πn/2, n∈Z x₂ = π × 1/3 + 2πn × 1/3
x₂ = π/3 + 2πn/3, n∈Z
ответ: x₁ = π/4 + πn/2, n∈Z
x₂ = π/3 + 2πn/3, n∈Z
