∵f(π/3)=4sin(π/3)+mcos(π/3-π/3)=2√3+m=0
∴m=-2√3 ∴f(x)=4sinx-2√3cos(x-π/3)=sinx-√3·cosx
=2(½sinx-√3/2·cosx)=2(sinπ/6sinx-cosπ6cosx)
=-2(cosπ/6cosx-sinπ/6sinx)=-2cos(x-π/6)
又∵2(½sinx-√3/2·cosx)=2(cosπ/3sinx-sinπ/3cosx)
=2sin(x-π/3)
∴f(x)=2sin(x-π/3)=-2cos(x-π/6)
把x=a+π/6代入上式得2sin(a+π/6-π/3)=8/5,-2cos(a+π/6-π/6)=8/5
化简得sin(a-π/6)=4/5,cosa=-4/5
又sin(a-π/6)=sinacosπ/6-cosasinπ/6=√3/2sina-1/2cosa=4/5
解得sina=4√3/15
sin(2a+π/6)=sin2acosπ/6+cos2asinπ/6
=√3sinacosa+cos²a-½
=√3×4√3/15+(-4/5)²-½
=47/50