解由2acosA=bcosC+ccosB
得2sinAcosA=sinBcosC+sinCcosB
即2sinAcosA=sin(B+C)
即2sinAcosA=sinA
即2cosA=1
即cosA=1/2,
即A=60°
2 有1知,B+C=120°
cosB-√3sinC=cos(120°-C)-√3sinC
=cos120°cosC+sin120°sinC-√3sinC
=-1/2cosC+√3/2sinC-√3sinC
=-1/2cosC-√3/2sinC
=-(sin30°cosC+cos30°sinC)
=-sin(C+30°)
由0°<C<120°
即30°<C+30°<150°
即1/2<sin(C+30°)<1
即-1<-sin(C+30°)<-1/2
即cosB-√3sinC的范围是(-1,-1/2)