设函数f(x)=2sin(2x-π/6)在x∈【0,π/2】上两个零点,则m的取值范围f(x)=2sin(2x-π/6)-m
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![设函数f(x)=2sin(2x-π/6)在x∈【0,π/2】上两个零点,则m的取值范围f(x)=2sin(2x-π/6)-m](/uploads/image/z/10378863-63-3.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D2sin%282x-%CF%80%2F6%29%E5%9C%A8x%E2%88%88%E3%80%900%2C%CF%80%2F2%E3%80%91%E4%B8%8A%E4%B8%A4%E4%B8%AA%E9%9B%B6%E7%82%B9%2C%E5%88%99m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4f%28x%29%3D2sin%282x-%CF%80%2F6%29-m)
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设函数f(x)=2sin(2x-π/6)在x∈【0,π/2】上两个零点,则m的取值范围f(x)=2sin(2x-π/6)-m
设函数f(x)=2sin(2x-π/6)在x∈【0,π/2】上两个零点,则m的取值范围
f(x)=2sin(2x-π/6)-m
设函数f(x)=2sin(2x-π/6)在x∈【0,π/2】上两个零点,则m的取值范围f(x)=2sin(2x-π/6)-m
f(x)=2sin(2x-π/6)里面没有m啊,你把题目改下吧
话说这题就是考函数的周期问题.简单啊.
m 在哪里?
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