As Andre mentioned in his comment, the more common terminology is "particular solution".
Your homogeneous solution $y_{hom} = C_1e^{2x} + C_2e^x$ a solution, not to the original equation, but to the homogeneous equation y'' - 3y' + 2y = 0, regardless of the constant parameters $C_1$ and $C_2$.
To find the particular (complementary) solution, we must consider solutions of the form $y_p=A\cos(wx) + B\sin(wx)\tag{1}$ After finding y_p'=wB\cos(wx) - wA\sin(wx) and y_p''=-w^2A\cos(wx) - w^2B\sin(wx) we substitute them into the original equation to get:
$-w^2A\cos(wx) - w^2Bsin(wx) - 3wB \cos(wx) + 3wA\sin(wx) + 2A\cos(wx) + 2Bs\in(wx) = \cos(wx)$
Because $\sin(wx)$ and $\cos(wx)$ are linearly independent, we know that the coefficients of $\cos(wx)$ on the LHS must equal the coefficient of $\cos(wx)$ on the RHS, and similarly for $\sin(wx)$ which gives us the system of equations:
$\begin {cases} (-w^2 + 2)A + (-3w)B = 1 \\ (3w)A + (-w^2 + 2)B = 0 \end{cases}$
Solving this system of equations gives us specific values for the coefficients $A$ and $B$:
$A = \frac{2-w^2}{w^4+5w^2+4}, B = \frac{-3w}{w^4+5 w^2+4}$
(provided that the denominator of those fractions is non-zero of course - also notice that that is equivalent to the condition that the determinant of the coefficient matrix for $A$ and $B$, in the system of equations given above, is non-zero.)
which now provides us with a particular (complimentary) solution to the original differential equation:
$y_p=\frac{2-w^2}{w^4+5w^2+4}cos(wx) + \frac{-3w}{w^4+5 w^2+4}sin(wx)$
And because, in general, (f + g)' = f' + g', we can see that
(y_{hom} + y_p)'' - 3(y_{hom} + y_p)' + 2(y_{hom} + y_p) = (y_{hom}'' - 3y_{hom}' + 2y_{hom}) + (y_p'' -3y_p' +2y_p) and then (y_{hom}'' - 3y_{hom}' + 2y_{hom}) + (y_p'' -3y_p' +2y_p) = 0 + cos(wx) = cos(wx) thus the general solution will be $y=y_{hom}+y_p=C_1e^{2x} + C_2e^x + \frac{2-w^2}{w^4+5w^2+4}\cos(wx) + \frac{-3w}{w^4+5 w^2+4}\sin(wx)$