f(x)=根号3sin(2x-π/6)+2sin^2(x-π/12)求单调递增区间

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/16 00:32:22

f(x)=√3sin(2x-π/6)+2sin²(x-π/12)
=√3sin(2x-π/6)+1-cos(2x-π/6)
=2[√3/2*sin(2x-π/6)-1/2*cos(2x-π/6)]+1
=2sin(2x-π/6-π/6)+1
=2sin(2x-π/3)+1
令2kπ-π/2≤2x-π/3≤2kπ+π/2,
得：kπ-π/12≤x≤kπ+5π/12
所以单调递增区间为[kπ-π/12,kπ+5π/12] (k∈Z)
再问：那递减区间呢
再答：令2kπ+π/2≤2x-π/3≤2kπ+3π/2，得：kπ+5π/12≤x≤kπ+11π/12 所以单调递增区间为[kπ+5π/12，kπ+11π/12] (k∈Z)

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