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 ?

Note that you can recognize the product rule of differentiation on the left side, $$(\sin(x)y(x))'=\sin(x)y'(x)+\cos(x)y(x),$$ which greatly simplifies the equation.
