f(x)=4(sin2π/3cosx-cos2π/3sinx)cosx
=(2√3cosx+2sinx)cosx
=2√3cos2x+2sinxcosx
=sin2x+2√3(1+cos2x)/2
=sin2x+√3cos2x+√3
=2sin(2x+π/3)+√3
所以最大值=2+√3
f(x)=4(sin2π/3cosx-cos2π/3sinx)cosx
=(2√3cosx+2sinx)cosx
=2√3cos2x+2sinxcosx
=sin2x+2√3(1+cos2x)/2
=sin2x+√3cos2x+√3
=2sin(2x+π/3)+√3
所以最大值=2+√3