Question

关于x的方程cos^2x-ksinx+2k+1=0有解,求k的取值范围

Answer 1

cos^2x-ksinx+2k+1=0

1-sin^2x-ksinx+2k+1=0

sin^2x+ksinx-2k-2=0

△ =k^2+4(2k+2)=k^2+8k+8≥0

k≥-4+2√2,或,k≤-4-2√2

x1,x2∈[-1,1]

|x1x2|=|-2k-2|≤1,-3/2≤k≤-1/2

|x1+x2|=|k|≤2,-2≤k≤2

所以,k的取值范围：[-4+2√2,-1/2]