题目有误.按原题,左=0
应该是cos(A-C)+cosB=3/2
cos(A-C)+cosB=3/2
cos(A-C)-cos(A+C)=3/2
2sinAsinC=3/2
sinAsinC=3/4
∵ b²= ac
∴ sin²B=sinAsinC(正弦定理)
sin²B=3/4
sinB>0
sinB=√3/2
所以 B=60°或120°
题目有误.按原题,左=0
应该是cos(A-C)+cosB=3/2
cos(A-C)+cosB=3/2
cos(A-C)-cos(A+C)=3/2
2sinAsinC=3/2
sinAsinC=3/4
∵ b²= ac
∴ sin²B=sinAsinC(正弦定理)
sin²B=3/4
sinB>0
sinB=√3/2
所以 B=60°或120°