2sin3x*cos3x=корень3/2

2sinα * cosα=sin2α

2sin3x*cos3x=√3/2

sin2 * (3x) = √3/2

sin6x=√3/2

6x = (-1) ^n * arcsin (√3/2) + 2πn, n∈Z

6x = (-1) ^n * (π/3) + 2πn, n∈Z

x = (-1) ^n * (π/3) : 6 + (2πn) : 6, n∈Z

x = (-1) ^n * (π/18 + (πn) : 3, n∈Z

ответ:x = (-1) ^n * (π/18 + (πn) : 3, n∈Z