2a=5π/3-2b
sin2a=sin(5π/3-2b)=sin(2π-2b-π/3)
=sin(2b+π/3)=sin2bcosπ/3+cos2bsinπ/3=sin2b/2+√3cos2b/2
y=2-sin2a-cos2b
=2-sin2b/2-√3cos2b/2-cos2b
=2-[sin2b+(√3+2)cos2b]/2
sin2b+(√3+2)cos2b=(√6+√2)sin(2b+θ)
tanθ=√3+2>√3θ∈(π/3,π/2)
2b+θ∈(π/3,3π/2)
sin(2b+θ)能够取到最大值1,不能取到最小值-1
y=2-[sin2b+(√3+2)cos2b]/2
=2-(√6+√2)sin(2b+θ)/2
能取到最小值(4-√6-√2)/2
没有最大值