It is known that the function f (x) = ACOS (Wx + φ) (a > 0, w > 0, - π / 2 ≤ φ ≤ π / 2) defined on R, and the difference between the maximum and the minimum is 4 The distance between the two lowest points is π, and all the symmetry centers of the function y = sin (2x + π / 3) image are on the symmetry axis of the image y = f (x) (1) The expression of finding f (x) (2) If f (X. / 2) = 3 / 2 (x ∈ [- π / 2, π / 2], find the value of COS (X. - π / 3) (3) Let vector a = (f (x - π / 6), 1), vector b = (1, cosx), X ∈ (0. π / 2), if vector a * vector B + 3 ≥. Constant holds, find the value range of real number M

It is known that the function f (x) = ACOS (Wx + φ) (a > 0, w > 0, - π / 2 ≤ φ ≤ π / 2) defined on R, and the difference between the maximum and the minimum is 4 The distance between the two lowest points is π, and all the symmetry centers of the function y = sin (2x + π / 3) image are on the symmetry axis of the image y = f (x) (1) The expression of finding f (x) (2) If f (X. / 2) = 3 / 2 (x ∈ [- π / 2, π / 2], find the value of COS (X. - π / 3) (3) Let vector a = (f (x - π / 6), 1), vector b = (1, cosx), X ∈ (0. π / 2), if vector a * vector B + 3 ≥. Constant holds, find the value range of real number M

1.）A-(-A)=4,A=2
The distance between two adjacent lowest points is π, that is, the period is π, so 2 π / w = π, w = 2
sin(2x+π/3）=cos（2x-π/6）
f(x)=2cos(2x+φ）
The difference between the symmetry axis and the symmetry center of y = cosx is π / 2
So 2x + φ - (2x - π / 6) = π / 2
So φ = π / 3
So f (x) = 2cos (2x + π / 3)
2.) f (X. / 2) = 2cos (X. + π / 3) = 3 / 2, cos (X. + π / 3) = 3 / 4
Sin (X. + π / 3) = + - (radical 7) / 4
cos（x.-π/3）=cos（x.+π/3 -.2π/3）
=cos（x.+π/3）cos（2π/3）+sin（x.+π/3)sin( -2.π/3）
=-1 / 2cos (X. + π / 3) - (radical 3) / 2 * sin (X. + π / 3)
=-3 / 8 + - (radical 21) / 8
Third, there are questions