方程sin^2x+2sinx+a=0有解,则a的取值范围为?
人气:405 ℃ 时间:2019-09-22 02:57:28
解答
-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
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