=cos2x-1+√3sin2x
=2sin(2x+π/6)-1,
它的增區(qū)間由（2k-1/2)π<2x+π/6<(2k+1/2)π,k∈Z確定,
各減π/6,（2k-2/3)π<2x<(2k+1/3)π,
各除以2,（k-1/3)π
∴sin(2C+π/6)=1,2C+π/6=π/2,C=π/6.ab=2√3.①
由余弦定理,1=a^+b^-2*2√3*√3/2,
∴a^+b^=7,
(a-b)^=7-4√3=（2-√3）^,a>b,
∴a-b=2-√3,與①聯(lián)立,解得a=2,b=√3.