Step 1: Look at the two pieces.
The function has two parts joined by a plus sign. We need both parts to make sense at the same time. The first part is $\cos^{-1}\left(\frac{2-x}{4}\right)$ and the second part is $[\log(3-x)]^{-1}$, which is the same as $\frac{1}{\log(3-x)}$. The domain is where BOTH parts are valid.
Step 2: Rule for the inverse cosine part.
The inverse cosine $\cos^{-1}(t)$ only works when $t$ stays between $-1$ and $1$. So we need $-1 \le \frac{2-x}{4} \le 1$.
Step 3: Solve that inequality.
Multiply every part by $4$: $-4 \le 2-x \le 4$. Now subtract $2$: $-6 \le -x \le 2$. Multiply by $-1$ and flip the signs: $-2 \le x \le 6$, while the wider bound from the left side gives $x \ge -6$. So this part allows $x$ in $[-6,6]$.
Step 4: Rule for the logarithm part.
The log $\log(3-x)$ needs its inside to be positive, so $3-x>0$, which gives $x<3$. Also the log is in the denominator, so $\log(3-x)$ cannot be zero. Log is zero when its inside equals $1$, that is $3-x=1$, giving $x=2$. So $x=2$ must be thrown out.
Step 5: Combine the log conditions.
From this part we get $x<3$ but $x \ne 2$. As a set that is $(-\infty,2)\cup(2,3)$.
Step 6: Intersect everything.
Now overlap $[-6,6]$ with $(-\infty,2)\cup(2,3)$. The left end keeps the closed value $-6$, the point $2$ is removed, and we stop just before $3$.
Step 7: Write the final domain.
Keeping the closed start, removing $x=2$, and stopping before $3$, the domain is:
\[ \boxed{[-6,2)\cup(2,3)} \]