Given the function f (x) = a * B, where a = (2sinwx, - 1), B = (2Sin (2pai / 3-wx), 1) w > 0, the abscissa of the intersection of the image of F (x) and the straight line y = - 2 is in a linear relationship An arithmetic sequence with tolerance Pai (1) Finding the analytic expression of F (x) (2) If a = root 3, B + C = 3, f (a) = 2, the area of triangle ABC is obtained

Given the function f (x) = a * B, where a = (2sinwx, - 1), B = (2Sin (2pai / 3-wx), 1) w > 0, the abscissa of the intersection of the image of F (x) and the straight line y = - 2 is in a linear relationship An arithmetic sequence with tolerance Pai (1) Finding the analytic expression of F (x) (2) If a = root 3, B + C = 3, f (a) = 2, the area of triangle ABC is obtained

f(x)=a*b=2sinwx*2sin(2π/3-wx)-1
=4sinwx(√3/2coswx+1/2sinwx)-1
=2√3sinwxcoswx+2sin²wx-1
=√3sin2wx+cos2wx
=2(√3/2sin2wx+1/2cos2wx)
=2sin(2wx+π/6)
∵ f (x) and the abscissa of the intersection point of the line y = - 2
The tolerance is Pai, and - 2 is the minimum value of the function
The minimum positive period of the function is π
2 π / (2W) = π, w = 1
∴f(x)=2sin(2x+π/6)
two
f（A）=2
When sin (2a + π / 6) = 1, a is the inner angle of the triangle
∴ 2A+π/6=π/2
∴A=π/6
∵a=√3,b+c=3
∴b^2+c^+2bc=9 (1)
From the cosine theorem, it is concluded that
3=b^2+c^2-2bc×√3/2
b^2+c^2-√3bc=3 (2)
(1)-(2):
(2+√3)bc=6
bc=6(2-√3)
The area of triangle ABC
S=1/2bcsinA
=1/2*6(2-√3)*1/2
=(6-3√3)/2