Example 1. Solve the equation $|\operatorname{tg} x+\operatorname{ctg} x|=\frac{4}{\sqrt{3}}$.
$$
\left.\pm \frac{5 \pi}{8}, k \in Z\right\} \text {. }
$$

Interpretation 1: The original equation is transformed into $\sin \angle x= \pm \frac{\sqrt{ } 3}{2}$.

Solving, we get $x=\frac{\sqrt{2}}{2} \pm \frac{\pi}{6} \cdot(k \in Z)$
Solution 2. After simplification, we get
$$
3 \operatorname{tg}^{4} x-10 \operatorname{tg}^{2} x+3=0 \text {. }
$$

From $\operatorname{tg} x= \pm \sqrt{3}$, we get $x_{1}=k \pi \pm \frac{\pi}{3} .(k \in Z)$ From $\operatorname{tg} x= \pm \frac{\sqrt{3}}{3}$, we get $x_{2}=k \pi \pm \frac{\pi}{6} \cdot(k \in Z)$
Also, $\frac{2 \operatorname{tg} x}{1+\operatorname{tg}^{2} x}= \pm \frac{\sqrt{3}}{2}$.

From (1) and (3), we know that (1) and (2) have the same solutions, so $x_{1}, x_{2}$ are equivalent to $x$. At this point, $\left\{x \left\lvert\, x=\frac{k \pi}{2} \pm \frac{\pi}{6}\right., k \in Z\right\}$ $=\left\{x \left\lvert\, x=k \pi \pm \frac{\pi}{3}\right.\right.$ or $\left.x=k \pi \pm \frac{\pi}{6}, k \in Z\right\}$.
Example 2. Solve the equation $\frac{\operatorname{tg} x}{\cos ^{2} 3 x}-\frac{\operatorname{tg} 3 x}{\cos ^{2} x}=0$.
Solution 1: The original equation is transformed into
$$
\frac{1}{\cos x \cdot \cos 3 x}\left(\frac{\sin x}{\cos 3 x}-\frac{\sin 3 x}{\cos x}\right)=0 \text {.(1) }
$$

Simplifying, we get
$$
\sin x \cdot \cos x-\sin 3 x \cdot \cos 3 x=0
$$

That is, $\sin 6 x-\sin 2 x=0$.
From $6 x=k \pi+(-1)^{k} \cdot 2 x$, we know
$$
x_{1}=\frac{k \pi}{2}, \quad x_{2}=\frac{k \pi}{4}+\frac{\pi}{8} .(k \in Z)
$$

Solution 2: From (1), we can also get
$$
\sin x \cdot \cos x\left(8 \cos ^{4} x-8 \cos ^{2} x+1\right)=0
$$

From $\sin x \cdot \cos x=0$, we get $x_{3}=\frac{k \pi}{2} .(k \in Z)$ From $8 \cos ^{4} x-8 \cos ^{2} x+1=0$, we get
$$
\cos ^{2} x=\frac{2 \pm \sqrt{ } 2}{4} .
$$
If $\cos ^{2} \frac{\pi}{8}=\sin ^{2} \frac{3 \pi}{8}=\frac{2+\sqrt{2}}{4}$,
$$
\cos ^{2} \frac{3 \pi}{8}=\sin ^{2} \frac{\pi}{8}=\frac{2-\sqrt{2}}{4} \text {. }
$$

Thus, ${ }_{4}=2 k \pi \pm \frac{\pi}{8}, 2 k \pi \pm \frac{7 \pi}{8}$,
$$
2 k \pi \pm \frac{3 \pi}{8}, 2 k \pi \pm \frac{5 \pi}{8} . \quad(k \in Z)
$$

If (3) is transformed into
$$
\begin{array}{l}
\sin 2 x \cdot\left(\cos ^{2} x-\sin ^{2} \frac{\pi}{8}\right) \\
\cdot\left(\cos ^{2} x-\sin ^{2} \frac{3 \pi}{8}\right)=0,
\end{array}
$$

We get $\sin 2 x \cdot \cos \left(2 x-\frac{\pi}{4}\right)$
$$
\text { - } \boldsymbol{c o s}\left(2 x-\frac{3 \pi}{4}\right)=0 \text {. }
$$

Using the product-to-sum formula, we know
$2 \sin 2 x \cdot \cos 4 x=0$.

From (2) and (4), we know that (1) and (3) have the same solutions, so $x_{1} 、 x_{2}$ are equivalent to $x_{3}, x_{4}$.
Removing the extraneous roots $x=k \pi+\frac{\pi}{2}$
$(k \in Z)$ from the equivalent solutions, the remaining solutions are still equivalent. Therefore, the solution set of the original equation is $\left\{x \mid x=k \pi\right.$ or $x=\frac{k \pi}{4}+\frac{\pi}{8}$,
$$
\begin{array}{l}
k \in Z\}=\left\{x \left\lvert\, x=2 k \pi \pm \frac{\pi}{8}\right. \text { or } x=2 k \pi\right. \\
\quad \pm \frac{7 \pi}{8} \text { or } x=2 k \pi \pm \frac{3 \pi}{8} \text { or } x=2 k \pi
\end{array}
$$