f(x)=mcos²x+√3msinxcosx+n
=m/2(1+cos2x)+√3/2msin2x+n
=m(√3/2sin2x+1/2cos2x)+m/2+n
=msin(2x+π/6)+m/2+n
∵x∈【0,π/6】.
∴2x+π/6∈[π/6,π/2]
∴1/2≤sin(2x+π/6)≤1
f(x)的值域为[3,4]
当m>0时,
f(x)max=3/2m+n=4
f(x)min=m+n=3
解得m=2,n=1
当m<0时,
f(x)min=3/2m+n=3
f(x)max=m+n=4
解得m=-2,n=6
(2)
m=2时,
f(x)=2sin(2x+π/6)+2
f(x)=2+√3,
sin(2x+π/6)=√3/2
∴2x+π/6=π/3,
x=π/12
当m=-2时,
f(x)=-2sin(2x+π/6)+5
f(x)=2+√3,
sin(2x+π/6)=(3-√3)/2
∴2x+π/6=arcsin(3-√3)/2
∴x=1/2arcsin(3-√3)/2-π/12