Verified answer

3sin2x - sin³x = 0

6*sinx*cosx - sin³x = 0

sinx*(6cosx - sin²x) = 0

sinx*(6cosx - (1-cos²x)) = 0

sinx*(cos²x + 6cosx - 1) = 0

1) sinx = 0

x = πn ,  n∈Z

2) cos²x + 6cosx - 1 = 0

a² + 6a - 1 = 0

D = 36 + 4 = 40

a = (- 6 ± 2√10)/2 = - 3 ± √10

a₁ = - 3 - √10 < - 3 - не подходит

a₂ = - 3 + √10 ≈ 0,16

cosx = √10 - 3

x = ± arccos(√10 - 3) + 2πk,  k∈Z

Ответ: πn ,  n∈Z ; ± arccos(√10 - 3) + 2πk,  k∈Z