求一道三角函数题已知A,B,C为△ABC的三个内角,a,b,c分别是三内角A,B,C的对边,方程(x平方-1)sinB-

(x^2-1)sinB-(x^2-x)sinC-(x-1)sinA=0
即(sinB-sinC)x^2+(sinC-sinA)x+(sinA-sinB)=0;
Δ=(sinC-sinA)^2-4(sinB-sinC)(sinA-sinB)
=(sinA)^2+4(sinB)^2+(sinC)^2-4sinAsinB+2sinAsinC-4sinBsinC
=(sinA-2sinB+sinC)^2
因为方程(x^2-1)sinB-(x^2-x)sinC-(x-1)sinA=0有两个相等实根,
所以Δ=0,则sinA-2sinB+sinC=0,
据正弦定理有a-2b+c=0,故a,b,c成等差数列.
tan(A/2)不能确定,其取值范围是:(0,√3/3).