corneliusy
幼苗
共回答了27个问题采纳率:85.2% 举报
1
f(x)=2sin(2x+60)-cos(2x+150)
=2[sin2x*cos60+sin60cos2x]-cos2xcos150sin2xsin150
=(3*sin2x/2)+(3√3cos2x/2)
当f(x)=(3√3)/2时
x=0或x=90
2
y=3/2cosx-f(x/2)-3√3sinx
=3/2cosx-【2[sinx*cos60+sin60cosx]-cosxcos150sinxsin150】-3√3sinx
=3/2cosx-【(3*sin2x/2)+(3√3cos2x/2)】-3√3sinx
=[(3-3√3)cosx/2]-[(6√3+3)sinx/2]
当x=π/2时,有最小值(6√3+3)/2.
当x=0时,有最大值(3-3√3)/2.
如有问题,
1年前
9