方程sin^2x+2sinx+a=0有解,则a的取值范围为?

题目

方程sin^2x+2sinx+a=0有解,则a的取值范围为?

答案

-a=sin²x+2sinx=(sinx+1)²-1
-1<=sinx<=1
0<=sinx+1<=2
所以0<=(sin+1)²<=4
所以-1<=(sin+1)²-1<=3
-1<=-a<=3
所以-3<=a<=1