设三角形ABC三内角满足于方程（sinB-sinA)x^2+(sinA-sinC)x=sinC-sinB=0有两相等的实数根,则三角形ABC中角B的取值范围是（）（sinB-sinA)x^2+(sinA-sinC)x+sinC-sinB=0

Question

设三角形ABC三内角满足于方程（sinB-sinA)x^2+(sinA-sinC)x=sinC-sinB=0有两相等的实数根,则三角形ABC中角B的取值范围是（ ）
（sinB-sinA)x^2+(sinA-sinC)x+sinC-sinB=0

Answer 1

(sinA-sinC)2-4（sinB-sinA)(sinC-sinB)
=sin2A-2sinAsinC+sin2C-4(sinBsinC-sinAsinC-sin2B+sinAsinB)
=(sinA+sinC)2-4sinB(sinA+sinC)+4sin2B=(sinA+sinC-2sinB)2
令上式=0得:2sinB=sinA+sinC=2sin[(A+C)/2]cos[(A-C)/2]
===>2sinB/2cosB/2=cosB/2cos[(A-C)/2]===>sinB/2=cos[(A-C)/2]/2
∵0≤(A-C)/2＜90o; ∴0＜cos[(A-C)/2]≤1
∴0＜sinB/2≤1/2,∴0o＜B/2≤30o ∴0o＜B≤60o