已知函数f(x)=2cosxsin(x+π3)−3sin2x+sinxcosx (Ⅰ)求函数f(x)的最小正周期; (Ⅱ)若x∈[−π2,π2]时,求f(x)的单调递减区间.

问题描述:

已知函数f(x)=2cosxsin(x+

π
3
)−
3
sin2x+sinxcosx
(Ⅰ)求函数f(x)的最小正周期;
(Ⅱ)若x∈[−
π
2
π
2
]
时,求f(x)的单调递减区间.

(Ⅰ)f(x)=2cosx(

1
2
sinx+
3
2
cosx)−
3
sin2x+sinxcosx
=2sinxcosx+
3
(cos2x−sin2x)

=sin2x+
3
cos2x

=2sin(2x+
π
3
)

∴T=π
(Ⅱ)f(x)的减区间为2kπ+
π
2
≤2x+
π
3
≤2kπ+
2
kπ+
π
12
≤x≤kπ+
12

又∵x∈[−
π
2
,−
π
12
]
,∴
π
2
≤x≤−
12
π
12
≤x≤
π
2

即f(x)在[−
π
2
,−
12
]
和在[
π
12
π
2
]
上单调递减.