设sinα,cosα是方程4x²-4mx+2m-1=0的两个根,且2分之3π

题目

设sinα,cosα是方程4x²-4mx+2m-1=0的两个根,且2分之3π

答案

sin^2a+cos^2a=(sina+cosa)^2-2sinacosa
=m^2-m+1/2
=1
m^2-m=1/2
(m-1/2)^2=3/4
m=1/2±√3/2
因为3π/2