Ответ:

sin(x) - cos(x) = (√2)*cos(3x). Разделим на -√2

1/√2 *cosx - 1/√2 *sinx = -cos(3x)

Пусть 1/√2 = cos(π/4)= sin(π/4)

cos(π/4)*cos(x)-sin(π/4)sin(x)=-cos(3x)

cos(x+π/4) +cos(3x) = 0

2*sin(2x+π/8)*sin(x-π/8) = 0

1) sin(2x+π/8) =0

2x+π/8 = πn, n є Z

x = -π/16 +πn/2, n є Z

2) sin(x-π/8) = 0

x-π/8 = πk, k є Z

x = π/8 + πk, k є Z

Ответ: π/8 + πk, k є Z ; x = -π/16 +πn/2, n є Z

Пошаговое объяснение: