Verified answer

(x^4)+(1/3)(x^2)sin(b)+(1/200)sin(╖/6)=0

or (x^4)+(1/3)(x^2)sin(b)+(1/200)*(1/2)=0

or (x^4)+(1/3)sin(b)(x^2)+(1/400)=0 Eqn A

As roots make an Arithmetic sequence, let the roots be (-3/2)d, (-1/2)d, (+1/2)d,(+3/2)d

The equation in terms of roots is:

{x+(3/2)d}*{x+(1/2)d}*{x-(1/2)d}*{x-(3/2)d} = 0

or {x^2-(9/4)d^2}*{x^2-(1/4)d^2} = 0

or x^4-(5/2)d^2*x^2+(9/16)d^4 = 0 Eqn B

Comparing the constant terms of Eqn A and Eqn B, we get

(9/16)d^4 = 1/400, and d^2 = 1/15

Comparing the coeffcients of x^2 of Eqn A and Eqn B, we get

(1/3)sin(b) = -(5/2)d^2 = -(5/2)*(1/15) = -1/6

sin(b) = -1/2

Therefore b = - ╖/6 or ╖ + ╖/6