If the solution of the differential equationdydx=1xcosy+sin2y is x=cesiny-k(1+siny), then the value of k is

dydx=1xcosy+2sinycosy​∴dxdy=xcosy+2sinycosy​dxdy+−cosyx=2sinycosy​I.F.=e−∫cosy dx=e−siny​∴ the solution is x.e−siny=2∫e−siny siny cosy dy​=−2sinye−siny+2∫e−sinycosydy​=−2sinye−siny−2e−siny+c​⇒x=−2siny−2+cesiny​=cesiny−21+siny+c​∴k=2