已知函数f（x）＝cos平方x－√3sinxcosx＋1

f(x)=cos²x -√3sinxcosx +1
=(1+cos2x)/2-(√3/2)sin2x +1
=sin(π/6 - 2x)＋3/2
因为,f(a)=5/6
所以,sin(π/6 -2a)＋3/2=5/6
即,sin(2a-π/6)＝2/3
因为,π/3≤a≤2π/3
所以,π/2≤2a-π/6≤7π/6
所以,cos(2a-π/6)＝-(√5)/3
sin2a＝sin[(2a-π/6)+(π/6)]
＝sin(2a-π/6)cos(π/6)+cos(2a-π/6)sin(π/6)
＝(2/3)×(√3/2)+(-√5/3)×(1/2)
＝(√3/3)-(√5/6)
＝(2√3 -√5)/6

f(x)=cos²x-√3sinxcosx＋1
=(1+cos2x)/2-(√3/2)sin2x+1
=1/2+(1/2)cos2x-(√3/2)sin2x+1
=3/2+sin(π/6)cos2x-cos(π/6)sin2x
=3/2+sin(π/6-2x)
=3/2-sin(2x-π/...