y=sin^2x+2sinx*cosx+3cos^2x

问题描述:

y=sin^2x+2sinx*cosx+3cos^2x
求最小值及此时x的值的集合

y=sin^2x+2sinx*cosx+3cos^2x
=sin2x+cos2x+2
=根号2sin(2x+pi/4)+2
所以最小值是2-根号2
次时
2x+pi/4=2kpi-pi/2
x=kpi-3pi/8