Question

方程2sin^2x-cosx+a-1=0有解,求a的范围

Answer 1

2sin²x-2-cosx+a+1=0

-2cos²x-cosx+a+1=0

2cos²x+cosx-a-1=0

令t=cosx,则：

2t²+t-a-1=0

关于t的二次方程有解，△=1-4*2*(-a-1)=8a+9>=0,a>=-9/8

-1<=-(a+1)/2<=1,-2<=a+1<=2, -3<=a<=1

所以： -9/8<=a<=1