已知函数y=sin²x+2sinxcosx+3cos²x,x∈R,问:

y=1+sin(2x)+2cos^2(x)
=1+sin(2x)+1+cos(2x)
=2+sin(2x)+cos(2x)
=2+√2sin(2x+π/4)
所以周期为π
(2) -π/2+2kπ<=2x+π/4<=π/2+2kπ
-3π/4+2kπ<=2x<=π/4+2kπ
-3π/8+kπ<=x<=π/8+kπ
(3)y=2+√2sin(2x+π/4) =2+√2sin(2(x+π/8))
所以是先左移π/8,然后上升2得到