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f(x)=(1-cos2x)/2+√3/2*sin2x
=√3/2sin2x-1/2*cos2x+1/2
=sin2xcosπ/6-cos2xsinπ/6+1/2
=sin(2x-π/6)+1/2
π/4<=x<=π/2
π/2<=2x<=π
π/3<=2x-π/6<=5π/6
所以2x-π/6=π/2时
原式最大=1+1/2=3/2
=√3/2sin2x-1/2*cos2x+1/2
=sin2xcosπ/6-cos2xsinπ/6+1/2
=sin(2x-π/6)+1/2
π/4<=x<=π/2
π/2<=2x<=π
π/3<=2x-π/6<=5π/6
所以2x-π/6=π/2时
原式最大=1+1/2=3/2
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