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.
解:(1)y=sin2x+2sinxcosx+3cos2x
=(sin2x+cos2x)+sin2x+2cos2x
=1+sin2x+(1+cos2x)
=sin2x+cos2x+2
=2