7. Evaluate the solution and the answer provided below. Indicate all errors and shortcomings.

Solve the equation $a\left(\frac{1}{\cos x}-\operatorname{tg} x\right)=1$.

Solution: Domain of definition: $x \neq \frac{\pi}{2}+\pi k, k \in \mathbb{Z}$. It is clear that when $a=0$ the equation
has no solutions. When $a \neq 0, x \neq \frac{\pi}{2}+\pi k, k \in \mathbb{Z}$, we have,

$$
\begin{gathered}
a\left(\frac{1}{\cos x}-\operatorname{tg} x\right)=1 \Longleftrightarrow a \sin x+\cos x=a \Longleftrightarrow \\
\Longleftrightarrow \frac{a}{\sqrt{a^{2}+1}} \sin x+\frac{1}{\sqrt{a^{2}+1}} \cos x=\frac{a}{\sqrt{a^{2}+1}} \Longleftrightarrow \\
\Longleftrightarrow \cos \left(x-\arccos \frac{1}{\sqrt{a^{2}+1}}\right)=\frac{a}{\sqrt{a^{2}+1}} \Longleftrightarrow \\
\quad \Longleftrightarrow x=\arccos \frac{1}{\sqrt{a^{2}+1}} \pm \arccos \frac{a}{\sqrt{a^{2}+1}}+2 \pi k, k \in \mathbb{Z}
\end{gathered}
$$

Answer: When $a \neq 0 x=\arccos \frac{1}{\sqrt{a^{2}+1}} \pm \arccos \frac{a}{\sqrt{a^{2}+1}}+2 \pi k, k \in \mathbb{Z}$; when $a=0$ the equation has no solutions.
1.II.7. Comment. The given solution contains two errors. Firstly, the transition from the equation $a \sin x+\cos x=a$ to the equation $\cos \left(x-\arccos \frac{1}{\sqrt{a^{2}+1}}\right)=\frac{a}{\sqrt{a^{2}+1}}$ is incorrect. The issue is that $\sin \left(\arccos \frac{1}{\sqrt{a^{2}+1}}\right)$ equals $\frac{a}{\sqrt{a^{2}+1}}$ or $-\frac{a}{\sqrt{a^{2}+1}}$ depending on the sign of the number $a$. Secondly, there was no verification of the solutions' belonging to the domain of the original equation. The equation $a \sin x+\cos x=a$ has solutions $x=\frac{\pi}{2}+2 \pi k$, which do not belong to the domain of the original equation.

Answer: $\arccos \frac{2 a}{a^{2}+1}+2 \pi k, k \in \mathbb{Z}$, for $|a| \geqslant 1 ;-\arccos \frac{2 a}{a^{2}+1}+2 \pi k$, $k \in \mathbb{Z}$, for $a \in(-1 ; 0) \cup(0 ; 1) ;$ for $a=0$ there are no solutions.