It helps knowing the independent solutions.

Just by looking at the quadradic it seems that e*sin(2t) is not among them.

Thus, by undetermined coefficients we look for a solution of the form:

x=A*sin(2t)+B*cos(2t)

x'=2A*cos(2t)-2B*sin(2t)

x''=-2A*sin(2t)-2B*cos(2t)

Thus,

[-2B-3A]sin(2t)+[2A-B]cos(2t)=e*sin(2t).

Thus,

-2B-3A=e

-B+2A=0

The solution to this system provides a particular solution.