已知向量
(I)求ω和常数a的值; (Ⅱ)求函数f(x)的单调递增区间. |
(I)f(x)=
•m
=2n
sinωxcosωx-2cos2ωx+a(1分)3
=
sin2ωx-cos2ωx-1+a=2sin(2ωx-3
)+a-1(3分)π 6
由T=
=π,得ω=1.(4分)2π 2ω
又当sin(2ωx-
)=1时ymax=2+a-1=3,得a=2(6分)π 6
(Ⅱ)由(I)知f(x)=2sin(2x-
)+1当2kπ-π 6
≤2x-π 2
≤2kπ+π 6
(k∈Z)(8分)π 2
即kπ-
≤x≤kπ+π 6
(10分)π 3
故f(x)的单调增区间为[kπ-
,kπ+π 6
],(k∈Z)(12分)π 3