Correct choice is (d) 3 cosx+2 siny=C
The best I can explain: Given that, \(\frac{dy}{dx}=\frac{3 \,sec\,y}{2cosec \,x}\)
\(\frac{2 \,dy}{sec \,y}=\frac{3dx}{cosec\,x}\)
Separating the variables, we get
2 cosy dy=3 sinx dx
Integrating both sides, we get
∫ 2 cosy dy = ∫ 3 sinx dx
2 siny=3(-cosx)+C
3 cosx+2 siny=C