设函数f(x)=√3sin(wx-π/3),且f( - 知识问答

设函数f(x)=√3sin(wx-π/3),且f(x)的周期为8,求f(x)的单调增区间

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解析：∵T=2π/│ω│=8,∴│ω│=2π/8=π/4
∴ ω＝±π/4
∴f(x)=√3sin(xπ/4-π/3)时,其单调递增区间为2kπ-π/2≤xπ/4-π/3≤2kπ+π/2,k∈Z
即8k-2/3≤x≤8k+10/3,k∈Z.
∴f(x)=√3sin(-xπ/4-π/3)时,其单调递增区间为2kπ-π/2≤-xπ/4-π/3≤2kπ+π/2,k∈Z
即-8k-10/3≤x≤-8k-2/3,k∈Z.

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