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 + sinAsinCcosAcosC - sinAsinC = - 1 / 2∴ cos(A+C) = - 1 / 2∵ A + C ∈ ( 0 ,π )∴ A + C = ( 2 / 3 )π∴ B = π - ( A + B ) = π / 3∴ sinB = ...