1. **State the problem:** Solve the differential equation $$\cos(x)\cos(y) \, dx + \sin(x)\sin(y) \, dy = 0$$. 2. **Separate variables:** Rewrite terms to isolate $dx$ and $dy$: $$\cos(x)\cos(y) \, dx = -\sin(x)\sin(y) \, dy$$ Divide both sides by $\sin(x)\cos(y)$ (assuming they are nonzero): $$\frac{\cos(x)}{\sin(x)} \, dx = -\frac{\sin(y)}{\cos(y)} \, dy$$ 3. **Integrate both sides:** $$\int \frac{\cos(x)}{\sin(x)} \, dx = -\int \frac{\sin(y)}{\cos(y)} \, dy$$ Note that $\frac{\cos(x)}{\sin(x)} = \cot(x)$ and $\frac{\sin(y)}{\cos(y)} = \tan(y)$. So integrals become: $$\int \cot(x) \, dx = -\int \tan(y) \, dy$$ 4. **Evaluate the integrals:** $$\int \cot(x) \, dx = \ln|\sin(x)| + C_1$$ $$\int \tan(y) \, dy = -\ln|\cos(y)| + C_2$$ Putting it together: $$\ln|\sin(x)| = \ln|\cos(y)| + C$$ where $C = C_2 - C_1$ is arbitrary constant. 5. **Solve for $y$:** Subtract $\ln|\cos(y)|$ from both sides: $$\ln\left| \frac{\sin(x)}{\cos(y)} \right| = C$$ Exponentiate both sides: $$\left| \frac{\sin(x)}{\cos(y)} \right| = e^{C} = A$$ Assuming $A > 0$, solve for $\cos(y)$: $$\cos(y) = \frac{\sin(x)}{A}$$ Take inverse cosine to get: $$y = \arccos \left( \frac{\sin(x)}{A} \right)$$ **Final answer:** $$y = \arccos \left( \frac{\sin(x)}{A} \right)$$ This represents the implicit general solution of the original differential equation, where $A$ is an arbitrary positive constant.