Particular solution for $\sin(x)y'(x)+y(x)\cos(x)=-\sin(2x)$. Let the ODE $$y'(x)\sin(x)+y(x)\cos(x)=-\sin(2x).$$
I know that a particular solution is of the form $w(x)=A\cos(2x)+B\sin(2x)$. When I put it in the equation, I get,
$$-2A\sin(2x)\sin(x)+2B\cos(2x)\sin(x)+A\cos(2x)\cos(x)+B\sin(2x)\cos(x)=-\sin(2x)$$
and thus $$\sin(2x)\Big(-2A\sin(x)+B\cos(x)\Big)+\cos(2x)\Big(2B\sin(x)+A\cos(x)\Big)=-\sin(2x),$$
and get the system
$$\begin{cases}
-2A\sin(x)+B\cos(x)=-1\\
2B\sin(x)+A\cos(x)=0.
\end{cases}.$$
How can I do now ?

Without RHS, the solution is
$$y_h=C\frac{1}{\sin(x)}.$$
a particular solution can be of the form
$$y_p=C(x)\frac{1}{\sin(x)}$$
with
$$C'(x)=-\sin(2x)$$
$\implies$
$$C(x)=\frac{1}{2}\cos(2x)$$
and
$$y_p=\frac{1}{2}\frac{\cos(2x)}{\sin(x)}$$
