y=√3 sinwxcoswx-cos^2wx+3/2

y=√3 sinwxcoswx-cos^2wx+3/2
=√3/2 sin2wx-(1+cos2wx)/2+3/2
=√3/2 sin2wx-1/2 cos2wx+1
=sin(2wx-z)+1
其中tanz=(1/2)/(√3/2)=√3/3
所以z=π/6

y=sin(2wx-π/6)+1
T=2π/|2w|=π
|w|=1
w=1或w=-1
x=π/6
y有最小值
所以2w*π/6-π/6=2kπ-π/2
w=6k-1
所以w=-1

所以y=sin(-2x-π/6)+1

sinx单调增则
2mπ-π/2<=x<=2mπ+π/2
所以
2mπ-π/2<=-2x-π/6<=2mπ+π/2
2mπ-π/3<=-2x<=2mπ+2π/3
-mπ-π/3<=x<=-mπ+π/6
即kπ-π/3<=x<=kπ+π/6

y可化为y=√3/2sin2wx-（1+cos2wx）/2+3/2

y=sin（2wx-π/6）+1

因为最小正周期π

所以2π/2w=π

w=1

y=sin（2x-π/6）+1

-π/2+2kπ≤2x-π/6≤π/2+2kπ

kπ-π/6≤x≤π/3+kπ

。。。题是不是给错了？还有个未知数应该

y=√3 sinwxcoswx-cos^2wx+3/2
y=√3/2sin2wx-（1+cos2wx）/2+3/2
y=sin（2wx-π/6）+1
最小正周期π
2π/2w=π
w=1
y=sin（2x-π/6）+1
单调递增区间
2x-π/6∈[-π/2+2nπ,π/2+2nπ]
当x=π/6时，函数有最小值2x-π/6=π/6