三角形ABC中1/2+2cosAcosC=cos(A-C),（1）a+c=4,三角形ABC的面积为(3根号3/4),求b

1 / 2 + 2cosAcosC = cos(A-C)
1 / 2 + 2cosAcosC = cosAcosC + sinAsinC
cosAcosC - sinAsinC = - 1 / 2
∴ cos(A+C) = - 1 / 2
∵ A + C ∈ ( 0 ,π )
∴ A + C = ( 2 / 3 )π
∴ B = π - ( A + B ) = π / 3
∴ sinB = √3 / 2
∵ S = 1 / 2 · ac · sinB = ( 3 / 4 )√3
∴ ac = 3
又∵ a + c = 4
解得a=3,c=1或a=1,c=3
由余弦定理 cosB = ( a² + c² - b² ) / 2ac
得 b = √7