已知函数f（x）=4cosxsin（x+π/6）- - 知识问答

已知函数f（x）=4cosxsin（x+π/6）-1求f（x）得单调区间

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答：
f(x)=4cosxsin(x+π/6)-1
=4cosx(sinxcosπ/6+cosxsinπ/6)-1
=2√3sinxcosx+2(cosx)^2-1
=√3sin2x+cos2x
=2*[(√3/2)sin2x+(1/2)cos2x]
=2sin(2x+π/6)
单调递增区间满足：2kπ-π/2

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