解由2B=A+C,A+B+C=180°
即B=60°
由 2b^2=3ac
即 2sinB^2=3sinAsinC
即 3sinAsinC=2*√3/2*√3/2
即sinAsinC=1/2
由sinAsinC=1/2[cos(A-C)-cos(A+C)]=1/2
即[cos(A-C)-cos(A+C)]=1
即cos(A-C)=1+cos(A+C)=1+cos120°=1+(-1/2)=1/2
即cos(A-C)=1/2=cos(±60°)
即A-C=±60°
又A+C=120
即A=90°或A=30°