По формуле sin α-sinβ=2sin(α-β)/2*cos(α+β)/2
sin(π/3+2x)-sin4x=0
2sin (π/3+2x-4x)/2*cos(π/3+2x+4x)/2=0
2sin(π/6-x)*cos(π/6+3x)=0 -2sin(x-π/6)*cos(3x+π/6)=0
sin(x-π/6)=0 или cos (3x+π/6)=0
x-π/6=πn 3x+π/6=π/2+πn
x=π/6+πn. n∈z 3x=π/2-π/6+πn
3x=π/3+πn. x=π/9+πn/3