1. **Problem Statement:** Find the general solution of the differential equation $$\cos(x) \cos(y) \, dx + \sin(x) \sin(y) \, dy = 0$$. 2. **Rewrite the equation:** We have $$\cos(x) \cos(y) \, dx + \sin(x) \sin(y) \, dy = 0.$$ Rearranged, $$\cos(x) \cos(y) \, dx = -\sin(x) \sin(y) \, dy.$$ Divide both sides by $\cos(x) \cos(y)$: $$dx = -\frac{\sin(x) \sin(y)}{\cos(x) \cos(y)} \, dy = -\tan(x) \tan(y) \, dy.$$ Then, $$\frac{dx}{dy} = -\tan(x) \tan(y).$$ 3. **Separate variables:** Rearrange: $$\frac{dx}{-\tan(x)} = \tan(y) \, dy.$$ This can be written as $$-\frac{dx}{\tan(x)} = \tan(y) \, dy.$$ 4. **Integrate each side:** Recall that $\tan(z) = \frac{\sin(z)}{\cos(z)}$ and the integral $$\int \frac{1}{\tan(z)} \, dz = \int \cot(z) \, dz = \ln|\sin(z)| + C.$$ Thus $$-\int \frac{1}{\tan(x)} \, dx = -\int \cot(x) \, dx = -\ln|\sin(x)| + C_1,$$ and $$\int \tan(y) \, dy = -\ln|\cos(y)| + C_2.$$ So $$-\ln|\sin(x)| = -\ln|\cos(y)| + C,$$ where $C = C_2 - C_1$. 5. **Simplify:** Multiply both sides by -1: $$\ln|\sin(x)| = \ln|\cos(y)| - C.$$ Exponentiate both sides: $$|\sin(x)| = e^{-C} |\cos(y)|.$$ Let $K = e^{-C} > 0$, then $$|\sin(x)| = K |\cos(y)|.$$ 6. **General solution:** This can be written as $$\sin(x) = \pm K \cos(y),$$ or equivalently $$\sin(x) = C \cos(y),$$ where $C$ is an arbitrary constant. **Final answer:** $$\boxed{\sin(x) = C \cos(y)}.$$