若关于x的方程4cosx-(cosx)^2+m-3=0恒有实数解,则实数m的取值范围是_______
人气:210 ℃ 时间:2019-11-04 12:17:36
解答
4cosx-(cosx)^2+m-3=0
-4+4cosx-(cosx)^2+m+1=0
m+1=(2-cosx)^2
f(x)=(2-cosx)^2的值域为[1,9]
所以m属于[0,8]
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