Substitution as suggested by the equation
sinxsinydx + cosxcosydy = 0
sinxsinydx+cosxcosydy=0\sin x \sin y dx+\cos x \cos y dy=0sinxsinydx+cosxcosydy=0
sinxcosxdx=−cosysinydy\frac{\sin x }{\cos x} dx=-\frac{\cos y}{ \sin y} dycosxsinxdx=−sinycosydy
−dcosxcosx=−dsinysiny-\frac{d\cos x }{\cos x}=-\frac{d\sin y}{ \sin y}−cosxdcosx=−sinydsiny
ln∣cosx∣=ln∣siny∣+lnC\ln|\cos x|=\ln|\sin y|+\ln Cln∣cosx∣=ln∣siny∣+lnC
cosx=Csiny\cos x=C\sin ycosx=Csiny
Answer: cosx=Csiny\cos x=C\sin ycosx=Csiny.
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