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
sin2x+sinxcosx
3
(Ⅰ)求函数f(x)的最小正周期;
(Ⅱ)若x∈[−
,π 2
]时,求f(x)的单调递减区间. π 2
答
(Ⅰ)f(x)=2cosx(
sinx+1 2
cosx)−
3
2
sin2x+sinxcosx
3
=2sinxcosx+
(cos2x−sin2x)
3
=sin2x+
cos2x
3
=2sin(2x+
)π 3
∴T=π
(Ⅱ)f(x)的减区间为2kπ+
≤2x+π 2
≤2kπ+π 3
,kπ+3π 2
≤x≤kπ+π 12
7π 12
又∵x∈[−
,−π 2
],∴−π 12
≤x≤−π 2
或5π 12
≤x≤π 12
π 2
即f(x)在[−
,−π 2
]和在[5π 12
,π 12
]上单调递减.π 2