Example 7. Solve the equation:
$$
\sin x + \cos x + \tan x = \sec x
$$
Solution: Using the universal substitution method: Let $\operatorname{tg} \frac{x}{2}=t$, then the original equation becomes: $\frac{2 t}{1+t^{2}}+\frac{1-t^{2}}{1+t^{2}}+\frac{2 t}{1-t^{2}}=\frac{1+t^{2}}{1-t^{2}}$. Eliminating the denominators and simplifying, we get $t(t-1)=0$.
$$
\begin{array}{l}
\therefore t_{1}=0, \quad t_{2}=1 . \\
\text { When } t_{1}=0, \text { i.e., } \operatorname{tg} \frac{x}{2}=0, \text { then } x_{1}=2 n \pi ;
\end{array}
$$

When $t_{2}=1$, i.e., $\operatorname{tg} \frac{x}{2}=1$, then $x_{2}=2 n \pi+\frac{\pi}{2}$.
Verification: Since $x=(2 n+1) \pi$ makes $\operatorname{tg} \frac{x}{2}$ undefined, we should consider possible lost roots:
$$
x=(2 n+1) \pi \text {. }
$$

Mark points $A(2 n \pi), B(2 n \pi$
$$
+\frac{\pi}{2} \text { ), } C(2 n \pi+\pi) \text { on the unit circle (Figure 14). }
$$

Comparing with the original equation, it is clear that $x=2 n \pi+\frac{\pi}{2}$ is an extraneous root, which should be discarded. And
$$
x=2 n \pi, \quad x=(2 n+1) \pi
$$

satisfy the original equation, so the solutions to the original equation are:
$$
x=k \pi \text {, ( } k \text { is an integer) }
$$