5. Answer: 1) $\alpha_{1}=\operatorname{arctg} \frac{24}{7}+2 \pi k, k \in Z \quad$ 2) $\alpha_{2}=\pi+2 \pi k, k \in Z$

$$
\text { 3) } \alpha_{3}=\operatorname{arctg} \frac{4+3 \sqrt{24}}{4 \sqrt{24}-3}+2 \pi k, k \in Z \quad \text { 4) } \alpha_{4}=\operatorname{arctg} \frac{3 \sqrt{24}-4}{4 \sqrt{24}+3}+2 \pi k, k \in Z
$$

(Other forms of the answers are possible: using other inverse trigonometric functions).
Solution (Geometric)

The line with the equation $x \cos a + y \sin a - 2 = 0$ is tangent to a circle of radius 2 centered at the origin. The system has a unique solution if it is tangent (common tangent) to the circle with the equation $x^{2} + y^{2} + 6x - 8y + 24 = 0 \leftrightarrow (x + 3)^{2} + (y - 4)^{2} = 1$, radius 1, and center at point $P(-3; 4)$. There are four cases of common tangents for two non-intersecting circles. Case 1.

![](https://cdn.mathpix.com/cropped/2024_05_06_706be2aaaec187922a3ag-13.jpg?height=454&width=531&top_left_y=2220&top_left_x=794)

Fig 1

In Fig 1, $L_{1}$ is the common tangent of the circles, $PQ$ is parallel to $L_{1}$, Triangle $OPQ$ is a right triangle with legs 3 and 4, equal to triangle $OPM$. The angle $POQ$ is $\beta, \operatorname{tg} \beta = 4 / 3$.

From the equality $\alpha + 2\beta = \pi$, we conclude that $\operatorname{tg} \alpha = -\operatorname{tg} 2\beta = -\frac{2 \operatorname{tg} \beta}{1 - \operatorname{tg}^{2} \beta} = \frac{24}{7}$, and the first series of solutions is $\alpha_{1} = \operatorname{arctg} \frac{24}{7} + 2\pi k, k \in \mathbb{Z}$.

Case 2. The second common tangent $L_{2}$ (see Fig. 1) is perpendicular to the x-axis and corresponds to the values of the parameter from the second series $\alpha_{2} = \pi + 2\pi k, k \in \mathbb{Z}$.

Case 3. The common tangent $L_{3}$ (does not intersect the segment $OP$)

![](https://cdn.mathpix.com/cropped/2024_05_06_706be2aaaec187922a3ag-14.jpg?height=568&width=557&top_left_y=687&top_left_x=727)

Fig 2

The angle $POM$ is $\beta, \operatorname{tg} \beta = 4 / 3$, the length of the segment $PQ = \sqrt{24}$, the angle $POQ$ is $\gamma, \operatorname{tg} \gamma = \sqrt{24}$.

From the equality $\alpha + \beta + \gamma = \pi$, it follows that $\operatorname{tg} \alpha = -\operatorname{tg}(\beta + \gamma) = -\frac{4 / 3 + \sqrt{24}}{1 - 4 \sqrt{24} / 3} = \frac{4 + 3 \sqrt{24}}{4 \sqrt{24} - 3}$.

This value corresponds to the third series of solutions: $\alpha_{3} = \operatorname{arctg} \frac{4 + 3 \sqrt{24}}{4 \sqrt{24} - 3} + 2\pi k, k \in \mathbb{Z}$.

Case 4. The common tangent $L_{4}$ is symmetric to the tangent $L_{3}$ with respect to the line $PO$, so (see Fig. 2) $\alpha_{4} = \alpha_{3} + 2\gamma, \operatorname{tg} \alpha_{4} = \frac{\operatorname{tg} \alpha_{3} + \operatorname{tg} 2\gamma}{1 - \operatorname{tg} \alpha_{3} \cdot \operatorname{tg} 2\gamma} = \frac{3 \sqrt{24} - 4}{4 \sqrt{24} + 3}$.

Then the fourth series of solutions is $\alpha_{4} = \operatorname{arctg} \frac{3 \sqrt{24} - 4}{4 \sqrt{24} + 3} + 2\pi k, k \in \mathbb{Z}$.