Question

Помогите срочно со 2 заданием ​

Answer 1

cos 2x•cosx - sin2x•sinx = 0

cos2x•cosx - 2cosx•sin²x = 0

cosx(cos2x-2sin²x)= 0

cosx(cos2x-1+cos2x)= 0

cosx(2cos2x-1) = 0

1)cos x = 0

x = π/2 + πn, n€Z

2)2cos2x = 1

cos2x = 1/2

2x = ±π/3 + 2πk

x = ±π/6 + πk, k€Z