1. **State the problem:** Solve the differential equation $$\sin y\,dy + \cos y(1 - \cos y \cos x)\,dx = 0.$$\n\n2. **Rewrite the equation:** The equation is $$\sin y\,dy + \cos y(1 - \cos y \cos x)\,dx = 0.$$ Rearranged as $$\sin y\,dy = -\cos y(1 - \cos y \cos x)\,dx.$$\n\n3. **Check if the equation is separable:** We want to write it as $$f(y) dy = g(x) dx.$$ Here, it is not immediately separable because of the mixed terms involving both $x$ and $y$ inside the parentheses.\n\n4. **Try to simplify the expression:** Expand the right side:\n$$-\cos y + \cos^2 y \cos x.$$\nSo the equation becomes:\n$$\sin y\,dy = -\cos y\,dx + \cos^2 y \cos x\,dx.$$\n\n5. **Rewrite as:**\n$$\sin y\,dy + \cos y\,dx = \cos^2 y \cos x\,dx.$$\n\n6. **Try substitution:** Let us consider $u = \sin y$. Then $du = \cos y\,dy$. But we have $\sin y\,dy$ and $\cos y$ terms, so this substitution is not straightforward.\n\n7. **Try to write the equation in differential form:**\n$$M(x,y) dx + N(x,y) dy = 0,$$ where\n$$M = \cos y (1 - \cos y \cos x), \quad N = \sin y.$$\n\n8. **Check if the equation is exact:**\nCalculate partial derivatives:\n$$\frac{\partial M}{\partial y} = -\sin y (1 - \cos y \cos x) + \cos y (\sin y \cos x) = -\sin y + \sin y \cos y \cos x + \cos y \sin y \cos x = -\sin y + 2 \sin y \cos y \cos x,$$\n$$\frac{\partial N}{\partial x} = 0.$$\nSince $$\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x},$$ the equation is not exact.\n\n9. **Try an integrating factor depending on $x$ or $y$:**\nCheck if $$\frac{1}{N} \left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \right)$$ depends only on $x$ or $y$.\nCalculate:\n$$\frac{1}{\sin y} (-\sin y + 2 \sin y \cos y \cos x - 0) = -1 + 2 \cos y \cos x.$$\nThis depends on both $x$ and $y$, so no integrating factor depending only on $x$.\n\nCheck $$\frac{1}{M} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) = \frac{0 - (-\sin y + 2 \sin y \cos y \cos x)}{\cos y (1 - \cos y \cos x)} = \frac{\sin y (1 - 2 \cos y \cos x)}{\cos y (1 - \cos y \cos x)}.$$\nDepends on both $x$ and $y$, so no integrating factor depending only on $y$.\n\n10. **Try substitution $z = \cos y$:**\nThen $dz = -\sin y dy$, so $\sin y dy = -dz$. The equation becomes:\n$$-dz + z(1 - z \cos x) dx = 0,$$\nor\n$$dz = z(1 - z \cos x) dx.$$\n\n11. **Rewrite as:**\n$$\frac{dz}{dx} = z(1 - z \cos x).$$\nThis is a separable differential equation.\n\n12. **Separate variables:**\n$$\frac{dz}{z(1 - z \cos x)} = dx.$$\n\n13. **Partial fraction decomposition:** Let $A$ and $B$ be functions of $x$ such that\n$$\frac{1}{z(1 - z \cos x)} = \frac{A}{z} + \frac{B}{1 - z \cos x}.$$\nMultiply both sides by $z(1 - z \cos x)$:\n$$1 = A(1 - z \cos x) + B z.$$\nEquate coefficients:\nConstant term: $1 = A$\nCoefficient of $z$: $0 = -A \cos x + B$\nSo $A = 1$, $B = \cos x$.\n\n14. **Rewrite integral:**\n$$\int \left( \frac{1}{z} + \frac{\cos x}{1 - z \cos x} \right) dz = \int dx.$$\n\n15. **Integrate with respect to $z$ (treating $x$ as constant):**\n$$\int \frac{1}{z} dz + \cos x \int \frac{1}{1 - z \cos x} dz = x + C.$$\n\n16. **Integrate each term:**\n$$\ln |z| - \ln |1 - z \cos x| = x + C,$$\nbecause $$\int \frac{1}{1 - a z} dz = -\frac{1}{a} \ln |1 - a z|,$$ here $a = \cos x$, so\n$$\cos x \times \left(-\frac{1}{\cos x} \ln |1 - z \cos x| \right) = -\ln |1 - z \cos x|.$$\n\n17. **Combine logarithms:**\n$$\ln \left| \frac{z}{1 - z \cos x} \right| = x + C.$$\n\n18. **Exponentiate both sides:**\n$$\left| \frac{z}{1 - z \cos x} \right| = e^{x + C} = K e^x,$$ where $K = e^C$.\n\n19. **Recall substitution:** $z = \cos y$, so\n$$\frac{\cos y}{1 - \cos y \cos x} = K e^x.$$\n\n20. **Final implicit solution:**\n$$\boxed{\frac{\cos y}{1 - \cos y \cos x} = C e^x},$$ where $C$ is an arbitrary constant.