y=sin²X+2sinX*cosX+3COS²X的最大值是?

解y=sin^2X+2sinX*cosX+3COS^2X
=sin^2X+COS^2X+2sinX*cosX+2COS^2X
=1+sin2x+2COS^2X
=2+sin2x+2COS&^2X-1
=2+sin2x+cos2x
=2+√2sin（2x+π/4)
≤2+√2
故
y=sin²X+2sinX*cosX+3COS²X的最大值是
2+√2

y=sin²x+2sinxcosx+3cos²x
=1+2cos²x+2sinxcosx
=1+cos2x+1+sin2x
=sin2x+cos2x+2
=√2sin(2x+π/4)+2
所以y的最大值为2+√2