f(x)=√3(1-cos2x)/2+1/2*sin2x
=sin2xcosπ/3-cos2xsinπ/3+√3/2
=sin(2x-π/3)+√3/2
2π/3<=2x-π/3<=5π/3
所以最大值是sin(2π/3)+√3/2=√3
最小是sin(3π/2)+√3/2=-1+√3/2
f(x)=√3(1-cos2x)/2+1/2*sin2x
=sin2xcosπ/3-cos2xsinπ/3+√3/2
=sin(2x-π/3)+√3/2
2π/3<=2x-π/3<=5π/3
所以最大值是sin(2π/3)+√3/2=√3
最小是sin(3π/2)+√3/2=-1+√3/2
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