设 三角形ABC的内角A,B,C所对的边a,b,c，若b^2=ac,cos(A-C)+cosB=3/2cos(A-C)+cosB=3/2=2・cos【（A-C+B

cos(A-C)+cosB=3/2
= 2・cos【（A-C+B）/2】・cos【（A-C-B）/2】 = 3/2 ,
即：2・cos【（π/2 - C）】・cos【（π/2 - A)】 = 3/2 ,
即：sinC・sinA = 3/4
b^2=ac , 由正弦定理：b^2/ac = 1 = (sinB)^2/sinAsinC = (sinB)^2/(3/4),
即：(sinB)^2 = 3/4 , sinB = 正负根3/2 ， B = 60°或120° ，
若B取120° ， 则cos(A-C) = 3/2 + 1/2 = 2 > 1,矛盾！，故B = 60°

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