在三角形ABC中,已知内角A= π/3,边BC=根号3,设内角B=x,周长为y.

（1）利用正弦定理a/sinA=b/sinB=c/sinC∴ √3/sin(π/3)=b/sinx=c/sin(2π/3-x)即 2=b/sinx=c/sin(2π/3-x)∴ b=2sinx,c=2sin(2π/3-x)∴ y=a+b+c=√3+2sinx+2sin(2π/3-x)=√3+2sinx+2sin(2π/3)cosx-2cos(2π/3)s...