Example 2. Express the following in terms of the inverse cosine:
(1) $\cos x=\frac{2}{5}, x \in\left(-\frac{\pi}{2}\right.$, );
(2) $\cos x=-\frac{\sqrt{2}}{3}, x \in\left(-\pi,-\frac{\pi}{2}\right)$.
Analysis: (1) $\because \cos x=\frac{2}{5}$,
$\therefore \quad \arccos \frac{2}{5} \in\left(0, \frac{\pi}{2}\right)$.
However, the problem is to find the angle within $\left(-\frac{\pi}{2}, 0\right)$ whose cosine value is $\frac{2}{5}$.

Since the cosine function is an even function (as shown in Figure 2), we have
$$
x=-\arccos \frac{2}{5}
$$
(2) $\because \cos x=-\frac{\sqrt{2}}{3}$,
Figure 2
$\therefore \quad \arccos \left(-\frac{\sqrt{2}}{3}\right) \in\left(\frac{\pi}{2}, \pi\right)$.
However, the problem is to find the angle within $\left(-\pi,-\frac{\pi}{2}\right)$ whose cosine value is $-\frac{\sqrt{2}}{3}$,

From Figure 3, we can see that $x=$ $-\arccos \left(-\frac{\sqrt{2}}{3}\right)$

When using the inverse cosine function to represent $x$ in $\cos x=h(|h|<1)$, if $x \in(-\pi, 0)$, then $x=-\arccos h$.

Similarly, we can also obtain $x$ in $\left(\frac{\pi}{2}, \frac{3}{2} \pi\right)$ for $\operatorname{tg} x=g$ and $x$ in $(-\pi, 0)$ for $\operatorname{ctg} x=f$:
$$
\begin{array}{l}
x=\pi+\operatorname{arctg} g, \\
x=-\pi+\operatorname{arcctg} f .
\end{array}
$$

Next, let's study the rules for representing angles using inverse trigonometric functions in any non-standard interval $[p, q]$. Generally, we have the following formulas:
If $x \in[p, q], p, q \in R$, and $q-p \leqslant \pi$, then
(1) When $\sin x=a,|a| \leqslant 1$,
$$
\begin{array}{l} 
x=p+\frac{\pi}{2}+\arcsin \\
\left\lfloor\sin \left(x-p-\frac{\pi}{2}\right)\right\rfloor ;
\end{array}
$$
(2) When $\cos x=a,|a| \leqslant 1$,
$$
x=p+\arccos [\cos (x-p)],
$$
(3) When $\operatorname{tg} x=a$,
$$
\begin{aligned}
x= & p+\frac{\pi}{2}+\operatorname{arctg} \\
& {\left[\operatorname{tg}\left(x-p-\frac{\pi}{2}\right)\right]_{3} }
\end{aligned}
$$
(4) When $\operatorname{ctg} x=a$,
$$
x=p+\operatorname{arcctg}[\operatorname{ctg}(x-p)] .
$$

Let's prove formula (1).
Proof: $\because x \in[p, q], p, q \in R$,
$$
\begin{array}{l}
\therefore \quad x-p-\frac{\pi}{2} \in\left[-\frac{\pi}{2}, q-p-\frac{\pi}{2}\right] . \\
\text { Also } q-p \leqslant \pi, \\
\therefore \quad q-p-\frac{\pi}{2} \leqslant \frac{\pi}{2} .
\end{array}
$$

Therefore, $\left[-\frac{\pi}{2}, q-p-\frac{\pi}{2}\right]$
$$
\subseteq\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] .
$$

Let $\sin \left(x-p-\frac{\pi}{2}\right)=a^{\prime}$, then
$$
\begin{array}{l}
x-p-\frac{\pi}{2}=\arcsin a^{\prime} \\
=\arcsin \left[\sin \left(x-p-\frac{\pi}{2}\right)\right],
\end{array}
$$

i.e., $x=p+\frac{\pi}{2}+\arcsin$
$$
\left[\sin \left(x-p-\frac{\pi}{2}\right)\right] \text {. }
$$

The proofs of formulas (2), (3), and (4) are omitted.
In the above formulas, $\sin x, \cos x, \operatorname{tg} x$, and $\operatorname{ctg} x$ are known, and $p$ is a constant. Therefore, $\sin \left(x-p-\frac{\pi}{2}\right), \cos (x-p)$, $\operatorname{tg}\left(x-p-\frac{\pi}{2}\right)$, and $\operatorname{ctg}(x-p)$ are all computable.