θ∈[0,π/2],且mcos²θ-msinθ+1-m>0恒成立,求实数m的取值范围
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θ∈[0,π/2],且mcos²θ-msinθ+1-m>0恒成立,求实数m的取值范围
θ∈[0,π/2],且mcos²θ-msinθ+1-m>0恒成立,求实数m的取值范围
θ∈[0,π/2],且mcos²θ-msinθ+1-m>0恒成立,求实数m的取值范围
cos²θ=1-sin²θ
mcos²θ-msinθ+1-m
=m(1-sin²θ)-msinθ+1-m>1
所以m(sin²θ+sinθ)
-msin²θ -msinθ +1 >0
m(sin²θ + sinθ)<1
m<1/(sin²θ+sinθ)
m<1/2