Example 3. Express $x$ in the form of the inverse cotangent in the following expression:
$$
\operatorname{tg} x=3, x \in(-3 \pi,-2 \pi) .
$$
Solve $p=-3 \pi$, from formula (4), we get
$x=p+\operatorname{arcctg}[\operatorname{ctg}(x-p)]$
$=-3 \pi+\operatorname{arcctg}[\operatorname{ctg}(-+3 \pi)]$
$=-3 \pi+\operatorname{arcctg}(\operatorname{ctg} x)$.
$\because \quad \operatorname{tg} x=3, x \in(\cdots 3 \pi, \quad-2 \pi)$,
$\therefore \operatorname{ctg} \approx=\frac{1}{3}$.
Thus, $\quad x=-3 \pi+\operatorname{arcctg} \frac{1}{3}$.