已知函数f(x)=3sin(x-ϕ)cos(x-ϕ)-cos2(x-ϕ)(0≤ϕ≤π2)为偶函数.(I)求函数的单调减区间;(II)把函数的图象向右平移π6个单位(纵坐标不变),得到函数g(x)的图象

题目简介

已知函数f(x)=3sin(x-ϕ)cos(x-ϕ)-cos2(x-ϕ)(0≤ϕ≤π2)为偶函数.(I)求函数的单调减区间;(II)把函数的图象向右平移π6个单位(纵坐标不变),得到函数g(x)的图象

题目详情

已知函数f(x)=
3
sin(x-ϕ)cos(x-ϕ)-cos2(x-ϕ)(0≤ϕ≤
π
2
)
为偶函数.
(I)求函数的单调减区间;
(II)把函数的图象向右平移
π
6
个单位(纵坐标不变),得到函数g(x)的图象,求方程g(x)+
1
2
=0
的解集.
题型:解答题难度:中档来源:不详

答案

(I)f(x)=
3
sin(x-ϕ)cos(x-ϕ)-cos2(x-ϕ)(0≤ϕ≤class="stub"π
2
)

=
3
2
sin2(x-φ)
-
1+cos2(x-φ)
2
=sin(2x-2φ-class="stub"π
6
)-class="stub"1
2

∵f(x)为偶函数,0≤ϕ≤class="stub"π
2
且,∴-2φ-class="stub"π
6
=class="stub"π
2
+kπ
,k∈Z,解得φ=class="stub"π
6

则f(x)=sin(2x-class="stub"π
2
)-class="stub"1
2
=-cos2x-class="stub"1
2

由2kπ-π≤2x≤2kπ(k∈Z)得,kπ-class="stub"π
2
≤x≤kπ,
故所求的递减区间是[kπ-class="stub"π
2
,kπ](k∈Z),
(II)函数的图象向右平移class="stub"π
6
个单位,得到函数g(x)的图象,则g(x)=-cos(2x-class="stub"π
3
-class="stub"1
2

由方程g(x)+class="stub"1
2
=0
得,-cos(2x-class="stub"π
3
)=0,即cos(2x-class="stub"π
3
)=0,解得2x-class="stub"π
3
=class="stub"π
2
+kπ
(k∈Z),
x=class="stub"5π
12
+class="stub"kπ
2
(k∈Z),
所求的解集为{x|x=class="stub"5π
12
+class="stub"kπ
2
(k∈Z)}.

更多内容推荐