1. The problem is to analyze the function $y = \log_{10}(\cos u)$.\n\n2. The logarithm function $\log_b(x)$ is defined only for $x > 0$ and $b > 0$, $b \neq 1$. Here, the base is 10, which is valid.\n\n3. The domain of $y = \log_{10}(\cos u)$ depends on where $\cos u > 0$. Since $\cos u$ oscillates between -1 and 1, the function is defined only where $\cos u > 0$.\n\n4. The cosine function is positive in intervals $(-\frac{\pi}{2} + 2k\pi, \frac{\pi}{2} + 2k\pi)$ for all integers $k$.\n\n5. The graph of $y = \log_{10}(\cos u)$ will have vertical asymptotes where $\cos u = 0$, i.e., at $u = \frac{\pi}{2} + k\pi$.\n\n6. The function decreases to $-\infty$ as $\cos u$ approaches 0 from the positive side, and increases without bound as $\cos u$ approaches 1 (where $\log_{10}(1) = 0$).\n\n7. Summary: The function is defined on intervals where $\cos u > 0$, has vertical asymptotes at zeros of $\cos u$, and the output is the logarithm base 10 of the cosine value.\n\nFinal answer: The domain is $\bigcup_{k \in \mathbb{Z}} \left(-\frac{\pi}{2} + 2k\pi, \frac{\pi}{2} + 2k\pi\right)$ and the function is $y = \log_{10}(\cos u)$ with vertical asymptotes at $u = \frac{\pi}{2} + k\pi$.