f(x)=1-cos2x-√3sin2x-1+√3
=√3-2sin(2x+π/6)
因为0≤x≤π/2
所以π/6≤2x+π/6≤7π/6
所以-1/2≤sin(2x+π/6)≤1
所以f(x)∈[√3-2,√3+1]零点是？sin(2x+π/6)=√3/22x+π/6=π/3或2π/3所以x=π/12或π/4不客气