已知函数f(x)=sin2x+23sinxcosx+3cos2x.(Ⅰ)求函数f(x)的最小正周期及单调递增区间;(Ⅱ)已知f(a)=3,且a∈(0,π),求a的值.-数学

题目简介

已知函数f(x)=sin2x+23sinxcosx+3cos2x.(Ⅰ)求函数f(x)的最小正周期及单调递增区间;(Ⅱ)已知f(a)=3,且a∈(0,π),求a的值.-数学

题目详情

已知函数f(x)=sin2x+2
3
sinxcosx+3cos2x.
(Ⅰ)求函数f(x)的最小正周期及单调递增区间;
(Ⅱ)已知f(a)=3,且a∈(0,π),求a的值.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)f(x)=sin2x+2
3
sinxcosx+3cos2x=
3
sin2x+2×class="stub"1+cos2x
2
+1=
3
sin2x+cos2x+2
=2sin(2x+class="stub"π
6
)+2.
所以最小正周期为:T=class="stub"2π
2

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

∴函数f(x)的单调增区间为 [-class="stub"π
3
+kπ,class="stub"π
6
+kπ]
 (k∈Z).
(Ⅱ)由f(a)=3,得2sin(2a+class="stub"π
6
)+2=3.
∴sin(2a+class="stub"π
6
)=class="stub"1
2

2a+class="stub"π
6
=class="stub"π
6
+2k1π
2a+class="stub"π
6
=class="stub"5π
6
+2k2π

即a=k1π或a=class="stub"π
3
+k2π

∵a∈(0,π),
∴a=class="stub"π
3

更多内容推荐