题目内容
已知
=(sinx,
cosx),
=(cosx,cosx),f(x)=
•
.
(1)若
⊥
,求x的取值集合;(2)求函数f(x)的周期及增区间.
| a |
| 3 |
| b |
| a |
| b |
(1)若
| a |
| b |
(1)∵
⊥
,∴
•
=0,
而
•
=sinxcosx+
cos2x=
sin2x+
cos2x+
=sin(2x+
)+
,
∴sin(2x+
)+
=0,即sin(2x+
)=-
,
∴2x+
=2kπ-
或2x+
=2kπ-
(k∈Z),
解得:x=kπ-
或x=kπ-
(k∈Z),
∴x的取值集合为{x|x=kπ-
或x=kπ-
(k∈Z)};
(2)∵f(x)=
•
=sin(2x+
)+
,∴f(x)的周期T=
=π,
∵y=sinx的增区间为[2kπ-
,2kπ+
](k∈Z),
由2kπ-
≤2x+
≤2kπ+
,解得:kπ-
≤x≤kπ+
,
∴f(x)的增区间为[kπ-
,kπ+
](k∈Z).
| a |
| b |
| a |
| b |
而
| a |
| b |
| 3 |
| 1 |
| 2 |
| ||
| 2 |
| ||
| 2 |
| π |
| 3 |
| ||
| 2 |
∴sin(2x+
| π |
| 3 |
| ||
| 2 |
| π |
| 3 |
| ||
| 2 |
∴2x+
| π |
| 3 |
| 2π |
| 3 |
| π |
| 3 |
| π |
| 3 |
解得:x=kπ-
| π |
| 2 |
| π |
| 3 |
∴x的取值集合为{x|x=kπ-
| π |
| 2 |
| π |
| 3 |
(2)∵f(x)=
| a |
| b |
| π |
| 3 |
| ||
| 2 |
| 2π |
| 2 |
∵y=sinx的增区间为[2kπ-
| π |
| 2 |
| π |
| 2 |
由2kπ-
| π |
| 2 |
| π |
| 3 |
| π |
| 2 |
| 5π |
| 12 |
| π |
| 12 |
∴f(x)的增区间为[kπ-
| 5π |
| 12 |
| π |
| 12 |
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