sin3x=-cos(x-п/6) необходимо полное решение

sin3x+cos(x-π/6)=0cos(π/2-3x)+cos(x-π/6)=02cos(π/6-x)cos(π/3-2x)=02sin(x+π/3)sin(π/6+2x)=0sin(x+π/3)=0x+π/3=πn,n∈zx=-π/3+πn,n∈zsin(π/6+2x)=0π/6+2x=πk,k∈z2x=-π/6+πk,k∈zx=-π/12+πk/2,k∈z