方程(sinB-sinA)x的平方+(sinA-sinC)x+sinC-sinB=0有两相等的实数根
则(sinA-sinC)²-4(sinB-sinA)(sinC-sinB)=0
由正弦定理化为角的形式
(a-c)²-4(b-a)(c-b)=0
(a+c-2b)²=0
所以a+c-2b=0 2b=a+c
由余弦定理cosB=(a²+c²-b²)/2ac
=[4a²+4c²-(a+c)²]/(8ac)
=(3a²+3c²-2ac)/(8ac)
=3(a²+c²)/(8ac)-1/4
≥3*2ac/(8ac)-1/4
=1/2
所以1/2≤cosB