System of differential equations with variable coefficients $$\mathbf{X}'(t) = \mathbf{A}(t)
\mathbf{X}(t),\  \mathbf{A}(t) = \begin{pmatrix}
2-2\sin t & \sin t\\ 
2-4\sin t & 1 + 2\sin t
\end{pmatrix}$$
We are asked to find the solution that satisfies that:
 $$\mathbf{X}_0 = \begin{pmatrix}
1 \\ 
4 
\end{pmatrix}$$
As a hint we are told to use:
$$\mathbf{X}(t) = \mathbf{P}(t)
\mathbf{Y}(t), \mathbf{P}(t) = \begin{pmatrix}
x_p & 0\\ 
y_p & 1
\end{pmatrix} $$

I would normally change the system using:
$$\mathbf{Y}'(t) = \mathbf{P}^{-1}(t)[\mathbf{A}(t)\mathbf{P}(t)-\mathbf{P}'(t)]
\mathbf{Y}(t)$$
But I don't know how to deal with the $x_p$ and $y_p$ in the matrix .Is there another way to tackle this problem?

First, by staring very hard at the system (or using a previous edit), we find a particular solution to the system, given by
\begin{equation}
\begin{pmatrix} 1 \\ 2 \end{pmatrix} e^{2 t}.
\end{equation}
Now, the notation 'subscript $p$' in the hint with $x_p$, $y_p$ is somewhat confusing, since we're clearly dealing with a periodic system here, with a particular solution. Turns out, the best thing to do is use the latter interpretation for '$p$'. In other words, when you use
\begin{equation}
x_p(t) = e^{2t},\quad y_p(t) = 2 e^{2t},
\end{equation}
the approach you proposed immediately leads to an explicitly solvable system.
