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已知函数f(x)=2根号3sinxcosx-2sin^2x+a,若f(x)有零点,求实数a的取值范围
人气:234 ℃ 时间:2019-09-20 05:41:54
解答
f(x)=√3sin2x+cos2x-1+a
=2sin(2x+π/6)-1+a=0有解
2sin(2x+π/6)=1-a
-2<=2sin(2x+π/6)<=2
所以-2<=1-a<=2
-1<=a<=3
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