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[函数] 存在性多参函数问题

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力工 Posted 2019-7-10 18:55 |Read mode
若存在实数$b$使得关于$x$的不等式$ |asin²x+(4a+b)sinx+13a+2b|-2sinx≤4$恒成立,
则实数$a$的取值为什么?
答案$[-1,1]$.

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facebooker Posted 2019-7-10 23:28
根号内a b合并同类项 外面的移项 两边同时除以2+sinx 后就很简单了

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