Sin3x=-cos(x-п/6) необходимо полное решение

Sin3x+cos(x-π/6)=0
cos(π/2-3x)+cos(x-π/6)=0
2cos(π/6-x)cos(π/3-2x)=0
2sin(x+π/3)sin(π/6+2x)=0
sin(x+π/3)=0
x+π/3=πn,n∈z
x=-π/3+πn,n∈z
sin(π/6+2x)=0
π/6+2x=πk,k∈z
2x=-π/6+πk,k∈z
x=-π/12+πk/2,k∈z