solve the ode $x^2y^\prime-1=\cos 2y$
First I wrote $y=y_h+y_p$. $y_h$ can be found easily i think: $$x^2y^\prime-1=0 \\y^\prime=\frac 1 x^2 \\ y=-\frac 1 x +c_1$$ but about finding the private solution i dont know from where to begin. suppose it was $\cos 2x$ i could have guessed $$y_p=a\cos 2x+ b\sin 2x$$ but here I can't do so. how can I find $y_p$?