2(√3/2сosx-1/2*sinx)-2cos4x=0
2cos(x+π/6)-2cos4x=0
cos(x+π/6)-cos4x=0
sin(-3x/2+π/12)sin(5x/2+π/12)=0
-sin(3x/2-π/12)*sin(5x/2+π/12)=0
sin(3x/2-π/12)=0⇒3x/2-π/12=πn⇒3x/2=π/12+πn⇒x=π/18+2πn/3,n∈z
sin(5x/2+π/12)=0⇒5x/2+π/12=πk⇒5x/2=-π/12+πk⇒x=-π/30+2πk/5,k∈z