write the given third order linear equation as an equivalent system of first order equations with initial values. Write the given third order linear equation as an equivalent system of first-order equations with initial values:
$$-(‴+3\sin(t)′) = 2t$$
with $$y(3)=−2,  y'(3)=3,  y''(3)=0$$
Use $_1=$, $_2=′$, and $_3=″$.
I was able to get $_3'$=$−(3\sin()_2)−2$. Does this mean that $x_1 = 0, x_3 =0$ but what is $x_2$ and the constant? What is the $t$?

You have that :
$$\begin {cases} x_1=y \\ x_2=y'=x'_1 \\x_3=y''=x'_2 \end{cases} \implies \begin {cases} x'_1=x_2 \\ x'_2=x_3 \end{cases}  $$
And as you noted you have also that 
$$x'_3=-3\sin(t)x_2-2t$$
Therefore you can write the DE equation as :
$$\begin{pmatrix} x_1 \\ x_2\\x_3 \end{pmatrix}'= \begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -3\sin(t) & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}+ \begin{pmatrix} 0 \\ 0 \\ -2t \end{pmatrix}$$
The "$t$" is the inhomogeneous part of the equation.
