$$
\begin{align}
u_2(t)&=\int\dfrac{y_1(t)g(t)}{ W[y_1(t),y_2(t)]}dt\\
\notag \\
&=\int\dfrac{\cos(2t) \cdot 3\csc(2t)}{2}dt\\
\notag \\
&=\dfrac{3}{2}\int\dfrac{\cos(2t)}{\sin(2t)}dt\\
\notag \\
&=\dfrac{3}{2}\int \cot(2t)dt\\
\notag \\
&=\dfrac{3}{4}\ln|\sin(2t)|\\
\end{align}
$$
Since,
$Y(t)=u_1(t)y_1(t)+u_2(t)y_2(t)$
Therefore,
$$
\begin{align}
Y(t)&= \cos(2t) \cdot (-\dfrac{3}{2}t)+\sin(2t) \cdot (\dfrac{3}{4}\ln|\sin(2t)|) \notag \\
\notag \\
&=\dfrac{3}{4}\sin(2t)\ln|\sin(2t)|-\dfrac{3}{2}t\cos(2t) \notag \\
\end{align}
$$
Thus, the general solution is,
$$
\begin{align}
y(t)&=y_c(t)+Y(t) \notag \\
\notag \\
y(t)&= c_1\cos(2t)+c_2\sin(2t)+ \dfrac{3}{4}\sin(2t)\ln|\sin(2t)|-\dfrac{3}{2}t\cos(2t) \notag \\
\end{align}
$$