求函数f(x)=sin(2x-π/3)-cos(2x+π/6)的最小正周期和单调递增区间

发布时间：2021-02-25 17:43:52

网友回答

利用cosa=sin(π/2-a)
f(x)=sin(2x-π/3)-cos(2x+π/6)
=sin(2x-π/3)-sin(-2x+π/3)
=sin(2x-π/3)+sin(2x-π/3)
=2sin(2x-π/3)
则T=π；由-π/2

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