A,B,C为三角形三内角,方程(sinB-sinA)x^2+(sinA-sinC)x+(sinC-sinB)=0有两个相等实数根,求B的范围

A,B,C为三角形三内角,方程(sinB-sinA)x^2+(sinA-sinC)x+(sinC-sinB)=0有两个相等实数根,求B的范围

sinA/a=sinB/b=sinC/c(其中a,b,c为角A,B,C对应的三条边)
设sinA/a=sinB/b=sinC/c=k
则a=sinA/k,b=sinB/k,c=sinC/k带入(sinB-sinA)x^2 +（sinA-sinC)x +(sinC-sinB)=0得
(b-a)x^2+(a-c)x+(c-b)=0
由有两相等实根得:
Δ=(a-c)^2-4(b-a)(c-b)=(2b-a-c)^2=0
所以2b=a+c即b=(a+c)/2(1)
a+b>c(2)
c+b>a(3)
(1)带入(2)得a>c/3(4)
(1)带入(3)得a<3c(5)
设a/c=y
则(4),(5)变为1/3<y<3(6)
(1)变为b/c=(y+1)/2(7)
cosB=(a^2+c^2-b^2)/(2ac)
=((a/c)^2+1-(b/c)^2)/(2a/c)
=(y^2+1-(y+1)^2/4)/(2y)
=3/8*(y+1/y)-1/4(8)
由y+1/y在(0,1)上递减,(1,+无穷)上递增,1/3<y<3得
y=1/3:y+1/y=10/3(取不到)
y=1:y+1/y=2(能取到)
y=3:y+1/y=10/3(取不到)
所以2<=y+1/y<10/3代入(8)
所以1/2<=cosB<1
所以0度<B<=60度

解： (1) 原式=(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≤π/3 (2) ∠B=π/3=60° cos[(A-C)/2]/2=1/2 A-C=0o A=C 即ΔABC的形状为等边三角形