已知f(x)=-cos2x+3sinxcosx.(1)求f(x)的最小正周期;(2)求f(x)的单调增区间.-数学

题目简介

已知f(x)=-cos2x+3sinxcosx.(1)求f(x)的最小正周期;(2)求f(x)的单调增区间.-数学

题目详情

已知f(x)=-cos2x+
3
sinxcosx

(1)求f(x)的最小正周期;
(2)求f(x)的单调增区间.
题型:解答题难度:中档来源:不详

答案

(1)∵f(x)=-class="stub"1+cos2x
2
+
3
2
sin2x=sin(2x-class="stub"π
6
)-class="stub"1
2
∴f(x)的最小正周期是π;
(2)令z=2x-class="stub"π
6
,函数y=sinz的单调增区间为[-class="stub"π
2
+2kπ,class="stub"π
2
+2kπ]

-class="stub"π
2
+2kπ≤2x-class="stub"π
6
≤class="stub"π
2
+2kπ
,得-class="stub"π
6
+kπ≤x≤class="stub"π
3
+kπ

y=sin(2x-class="stub"π
6
)-class="stub"1
2
的单调增区间为[-class="stub"π
6
+kπ,class="stub"π
3
+kπ](k∈Z)

更多内容推荐