Question

高中数学求解要过程！求15 16

Answer 1

15、
原式==> x^2*(x^2-2x-3)＜0
==> x≠0，且x^2-2x-3＜0
==> x≠0，且(x+1)(x-3)＜0
==> x≠0，且-1＜x＜3

16、
已知有两个相等的实数根，则：△=(sinA-sinC)^2-4(sinB-sinA)*(sinC-sinB)=0
==> sin^2 A+sin^2 C-2sinAsinC-4sinBsinC+4sin^2 B+4sinAsinC-4sinAsinB=0
==> sin^2 A+sin^2 C+2sinAsinC-4sinB(sinA+sinC)+4sin^2 B=0
==> (sinA+sinC)^2-4sinB(sinA+sinC)+4sin^2 B=0
==> [(sinA+sinC)-2sinB]^2=0
==> sinA+sinC=2sinB
==> 2sin(A+C/2)cos(A-C/2)=2*2sin(B/2)cos(B/2)
==> 2cos(B/2)cos(A-C/2)=4sin(B/2)cos(B/2)
==> sin(B/2)=(1/2)cos(A-C/2)
因为A-C∈(-π，π)，则(A-C)/2∈(-π/2,π/2)
那么，cos(A-C/2)∈(0,1)
所以，sin(B/2)=(1/2)cos(A-C/2)∈(0,1/2)
则，B/2∈(0,π/3)
那么，B∈(0,2π/3)