Sinx-√2 sin3x=-sin5x cos(70°+x)cos(x-20°)=1/2 - вопрос №2357020125851452 от saitieva03 25.04.2021 19:07

Question

Алгебра: Sinx-√2 sin3x=-sin5x cos(70°+x)cos(x-20°)=1/2

saitieva03 · Answer

1) Sinx - √2Sin3x + Sin5x = 0 2Sin3xCos2x - √2Sin3x = 0 Sin3x(Cos2x -√2) = 0 Sin3x = 0 или Cos2x -√2= 0 3x = πn , n∈Z Cos2x =√2 x = πn/3, n ∈Z нет решений. 2) Сos(70°+x)Cos(x -20°) = 1/2 Sin(20° -x)Cos(20° -x) = 1/2 |*2 2Sin(20° -x)Cos(20° -x) = 1 Sin(40° -2x) = 1 40° -2x = 90° + 360°*n, n ∈Z 2x = -50° -360°n, n∈Z x = -25° -180°n, n ∈Z...