f(x)=2a·b-1=2√3sinxcosx+2cos²x-1
=√3sin2x+cos2x
=2sin(2x+π/6)
函数f（x）的单调增区间
2x+π/6∈[(2k-1/2)π,(2k+1/2)π]
x∈[(k-1/3)π,(k+1/6)π]
若a平行b ,a×b=0,√3cos²x-sinxcosx=0,tanx=√3,或不存在
若a垂直b,a·b=0,√3sinxcosx+cos²x=0
2cosxsin(x+π/6)=0
x=(k+1/2)π或(k-1/6)π
x最小正值
π/2