cosx^2+pcosx-p^2+1=0恒有解,求实数P的范围

首先判别式大于等于0
p^2+4p^2-4>=0
p^2>=4/5
p>=2√5/5,p<=-2√5/5

cosx=[-p±√(5p^2-4)]/2
-1<=cosx<=1
-1<=[-p±√(5p^2-4)]/2<=1
-2<=-p±√(5p^2-4)<=2
p-2<=±√(5p^2-4)<=p+2

若p-2<=√(5p^2-4)<=p+2
p-2<=√(5p^2-4)
p<=2成立
p>2
p^2-4p+4<=5p^2-4
p>=1,p<=-2,则p>2
所以恒成立
√(5p^2-4)<=p+2
p<=-2不成立
p>-2
5p^2-4<=p^2+4p+4
-1<=p<=2
所以-1<=p<=2

若p-2<=-√(5p^2-4)<=p+2
-p-2<=√(5p^2-4)<=-p+2
-p-2<=√(5p^2-4)
p>=-2成立
p<-2
p^2+4p+4<=5p^2-4
p>=2,p<=-1,则p<-2
所以恒成立
√(5p^2-4)<=-p+2
p>=2不成立
p<2
5p^2-4<=p^2-4p+4
-2<=p<=1
所以-2<=p<=1

综上-1<=p<=-2√5/5,2√5/5<=p<=1

根的分布，就是麻烦
因为-1<=cosx<=1，函数开口向上 所以要使它恒有解有四个条件：
（为了方便就把方程左边写成f（cosx））

1.判定式大于0（废话）
2.f(-1)&