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