Example 4. Express $x$ in the following equation using the inverse cosine function: $\cos x=\frac{3}{5} \quad x \in(3,6.1)$.
Here $p=3$, by formula (2), we get
$$
\begin{array}{c}
x=p+\arccos [\cos (x-p)] \\
=3+\arccos [\cos (x-3)] . \\
\because \quad \cos x=\frac{3}{5}, x \in(3,6.1), \\
\therefore \quad \sin x=-\frac{4}{5}, \\
\cos (x-3)=\frac{3}{5} \cos 3-\frac{4}{5} \sin 3 .
\end{array}
$$

Thus, $x=3+\arccos \left(\frac{3}{5} \cos 3\right.$
$$
\left.-\frac{4}{5} \sin 3\right) \text {. }
$$

In the above formula, it is always assumed that the interval $[p, q]$ satisfies $q-p \leqslant \pi$. When $q-p>\pi$, we only need to divide the interval $[p, q]$ into several smaller intervals $\left[p_{0}, p_{1}\right],\left[p_{1}, p_{2}\right], \cdots,\left[p_{n-1}, p_{n}\right]$, where $0<p_{i}-p_{i-1} \leqslant \pi(i=1,2, \cdots, n)$, $p_{0}=p, p_{n}=q$, and then express $x$ according to the formula in each of these $n$ subintervals.