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.
答
(1)y=sin2x+2sinxcosx+3cos2x
=(sin2x+cos2x)+sin2x+2cos2x
=1+sin2x+(1+cos2x)
=sin2x+cos2x+2
=
sin(2x+
2
)+2,π 4
∴函数的最小正周期T=
=π.2π 2
(2)由2kπ−
≤2x+π 2
≤2kπ+π 4
,得kπ−π 2
≤x≤kπ+3π 8
(k∈Z),π 8
∴函数的增区间为[kπ−
,kπ+3π 8
](k∈Z).π 8