= sin(3x/4)* sin(π+π/2+3x/4)* sin(2π+π/2+3x/2)

= sin(3x/4)*[-sin(π/2+3x/4)]* sin(π/2+3x/2)

=-sin(3x/4)*cos(3x/4)*cos(3x/2)

=(- 1/2)* sin(3x/2)* cos(3x/2)

=(- 1/4)*sin3x

( 1)∫- 1/4≤f(x)≤ 1/4

∴f(x) is the value of the abscissa intersecting with straight lines y=- 1/4 and y= 1/4.

F(x) obtains the abscissa corresponding to the maximum value.

X∈(0, ﹢∞), arranged in sequence {an} from small to large.

Then the sequence {an} is a arithmetic progression with a tolerance of f(x).

∴ Tolerance d=( 1/2)*(2π/3)=π/3

And when x=π/6, f (π/6) = (-1/4) * sin (3× π/6) =-1/4.

∴a 1=π/6

General term formula of ∴ sequence {an}

an=a 1+(n- 1)d

=π/6+(n- 1)*(π/3)

=-π/6+πn/3 (n∈N*)