已知函数y=sin2x+2sinxcosx+3cos2x,x∈R. (1)函数y的最小正周期; (2)函数y的递增区间.

问题描述:

已知函数y=sin2x+2sinxcosx+3cos2x,x∈R.
(1)函数y的最小正周期;
(2)函数y的递增区间.

(1)y=sin2x+2sinxcosx+3cos2x
=(sin2x+cos2x)+sin2x+2cos2x
=1+sin2x+(1+cos2x)
=sin2x+cos2x+2
=

2
sin(2x+
π
4
)+2,
∴函数的最小正周期T=
2
=π.
(2)由2kπ−
π
2
≤2x+
π
4
≤2kπ+
π
2
,得kπ−
8
≤x≤kπ+
π
8
(k∈Z),
∴函数的增区间为[kπ−
8
,kπ+
π
8
]
(k∈Z).