已知函数f(x)=3sin(2x-π6)+2sin2(x-π12)(x∈R).(I)求函数f(x)的最小正周期;(II)求函数f(x)的单调增区间.-数学

题目简介

已知函数f(x)=3sin(2x-π6)+2sin2(x-π12)(x∈R).(I)求函数f(x)的最小正周期;(II)求函数f(x)的单调增区间.-数学

题目详情

已知函数f(x)=
3
sin(2x-
π
6
)+2sin2(x-
π
12
)(x∈R)

(I)求函数f(x)的最小正周期;
(II)求函数f(x)的单调增区间.
题型:解答题难度:中档来源:不详

答案

(Ⅰ) f(x)=
3
sin(2x-class="stub"π
6
)+1-cos2(x-class="stub"π
12

=2[
3
2
sin2(x-class="stub"π
12
)-class="stub"1
2
 cos2(x-class="stub"π
12
)]+1=2sin[2(x-class="stub"π
12
)-class="stub"π
6
]+1
=2sin(2x-class="stub"π
3
)+1  
∴T=class="stub"2π
2

(Ⅱ)令2kπ-class="stub"π
2
≤2x-class="stub"π
3
≤2kπ+class="stub"π
2

解得kπ-class="stub"π
12
≤x≤kπ+class="stub"5π
12
,k∈Z

即函数的递增区间为:[kπ-class="stub"π
12
,kπ+class="stub"5π
12
],k∈Z

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