Помогите решить уравнение

1) sinx-√2sin3x=-sin5x

2) cos (70°+x) cos (x-20°) = 1/2

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