设△ABC三内角满足的方程（sinB-sinA)x^2+(sinA-sinC)x+(sinC-sinB)=0有两个相等的实根.

题目

设△ABC三内角满足的方程（sinB-sinA)x^2+(sinA-sinC)x+(sinC-sinB)=0有两个相等的实根.
（1）求证：角B不大于π/3
（2）当角B取最大值时,判断△ABC的形状
过程详细~~

答案

(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
(2)
∠B=60º:
cos[(A-C)/2]/2=1/2===>A-C=0º===>A=C
∴ΔABC为等边三角形