(2013•石景山区一模)已知函数f(x)=sin(2x+π6)+cos2x.
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(2013•石景山区一模)已知函数f(x)=sin(2x+
)+cos2x
π |
6 |
(Ⅰ)f(x)=sin(2x+
π
6)+cos2x=sin2xcos
π
6+cos2xsin
π
6+cos2x
=
3
2sin2x+
3
2cos2x=
3(
1
2sin2x+
3
2cos2x)=
3sin(2x+
π
3).
令 2kπ-
π
2≤2x+
π
3≤2kπ+
π
2,k∈z,求得 kπ-
5π
12≤x≤kπ+
π
12,
函数f(x)的单调递增区间为[kπ-
5π
12,kπ+
π
12],k∈z.
(Ⅱ)由已知f(A)=
3
2,可得 sin(2A+
π
3)=
π
6)+cos2x=sin2xcos
π
6+cos2xsin
π
6+cos2x
=
3
2sin2x+
3
2cos2x=
3(
1
2sin2x+
3
2cos2x)=
3sin(2x+
π
3).
令 2kπ-
π
2≤2x+
π
3≤2kπ+
π
2,k∈z,求得 kπ-
5π
12≤x≤kπ+
π
12,
函数f(x)的单调递增区间为[kπ-
5π
12,kπ+
π
12],k∈z.
(Ⅱ)由已知f(A)=
3
2,可得 sin(2A+
π
3)=
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