Question

. sin x cos ydx + cos x sin ydy = 0

Answer 1

$\sin x\cos ydx+\cos x\sin ydy=0\Leftrightarrow \sin x\cos y+y'\cos x\sin y=0\\y'=-\mathrm{ctg}y\cdot \mathrm{tg}x\Leftrightarrow \int y'\mathrm{tg}ydx=-\int \mathrm{tg}dx\\\ln \cos x+C=-\ln \cos y\Rightarrow y=\pm \arccos \cfrac{C}{\cos x}$