y = (√3tanx+1)cos²x = √3sinxcosx + cos²x = √3/2 sin2x + 1/2 cos 2x + 1/2
= cos 30 sin2x + sin 30 cos2x +1/2 = sin(2x+30) +1/2
故最大值为 1+ 1/2 = 3/2
y = (√3tanx+1)cos²x = √3sinxcosx + cos²x = √3/2 sin2x + 1/2 cos 2x + 1/2
= cos 30 sin2x + sin 30 cos2x +1/2 = sin(2x+30) +1/2
故最大值为 1+ 1/2 = 3/2