已知sinQ,cosQ是方程4x^2-4mx+2m-1=0的两根,3π/2<Q<2π,求角Q
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![已知sinQ,cosQ是方程4x^2-4mx+2m-1=0的两根,3π/2<Q<2π,求角Q](/uploads/image/z/8430680-56-0.jpg?t=%E5%B7%B2%E7%9F%A5sinQ%2CcosQ%E6%98%AF%E6%96%B9%E7%A8%8B4x%5E2-4mx%2B2m-1%3D0%E7%9A%84%E4%B8%A4%E6%A0%B9%2C3%CF%80%2F2%EF%BC%9CQ%EF%BC%9C2%CF%80%2C%E6%B1%82%E8%A7%92Q)
已知sinQ,cosQ是方程4x^2-4mx+2m-1=0的两根,3π/2<Q<2π,求角Q
已知sinQ,cosQ是方程4x^2-4mx+2m-1=0的两根,3π/2<Q<2π,求角Q
已知sinQ,cosQ是方程4x^2-4mx+2m-1=0的两根,3π/2<Q<2π,求角Q
x1+x2=-b/a=-(-4m)/4=m=sinQ+cosQ.
x1*x2=c/a=(2m-1)/4=sinQ*cosQ.
(sinQ+cosQ)^2=sinQ^2+cosQ^2+2sinQ*cosQ
即[(2m-1)/4]^2=1+2m
解得m=.
∵3π/2