设函数f(x)=2cos2x+3sin2x.(1)求f(x)的周期以及单调增区间;(2)当f(x)=53(-π6<x<π6)时,求sin2x.-数学

题目简介

设函数f(x)=2cos2x+3sin2x.(1)求f(x)的周期以及单调增区间;(2)当f(x)=53(-π6<x<π6)时,求sin2x.-数学

题目详情

设函数f(x)=2cos2x+
3
sin2x

(1)求f(x)的周期以及单调增区间;
(2)当f(x)=
5
3
(-
π
6
<x<
π
6
)
时,求sin2x.
题型:解答题难度:中档来源:不详

答案

(1)f(x)=2cos2x+
3
sin2x
=1-cos2x+
3
sin2x=2sin(2x+class="stub"π
6
)+1
∴函数f(x)的最小正周期T=class="stub"2π
2
=π,
当2kπ-class="stub"π
2
≤2x+class="stub"π
6
≤2kπ+class="stub"π
2
,即kπ-class="stub"π
3
≤x≤kπ+class="stub"π
6

故函数的单调增区间为[kπ-class="stub"π
3
,kπ+class="stub"π
6
](k∈Z)
(2)∵f(x)=2sin(2x+class="stub"π
6
)+1=class="stub"5
3

∴sin(2x+class="stub"π
6
)=class="stub"1
3
,且-class="stub"π
6
<x<class="stub"π
6

∴cos(2x+class="stub"π
6
)>0
∴cos(2x+class="stub"π
6
)=
1-class="stub"1
9
=
2
2
3

sin2x=sin(2x+class="stub"π
6
-class="stub"π
6
)=class="stub"1
3
×
3
2
-
2
2
3
×class="stub"1
2
=
3
-2
2
6

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