8.393. Find the angles $\alpha, \beta$ and $\gamma$ of the first quadrant, if it is known that they form an arithmetic progression with a common difference of $\frac{\pi}{12}$, and their tangents form a geometric progression
Solution.

Let $\alpha, \beta$ and $\gamma$ be members of an arithmetic progression, $d=\frac{\pi}{12}$; $\operatorname{tg} \alpha, \operatorname{tg} \beta, \operatorname{tg} \gamma$ be members of a geometric progression; $\beta=\alpha+\frac{\pi}{12}, \gamma=\alpha+\frac{\pi}{6}$.

By the property of members of a geometric progression

$$
b_{k}^{2}=b_{k-1} b_{k+1}, k=2,3, \ldots, n-1, \text { therefore } \operatorname{tg}^{2} \beta=\operatorname{tg} \alpha \operatorname{tg} \gamma,
$$

$$
\operatorname{tg}^{2}\left(\alpha+\frac{\pi}{12}\right)=\operatorname{tg} \alpha \operatorname{tg}\left(\alpha+\frac{\pi}{6}\right) \Leftrightarrow \operatorname{tg}^{2}\left(\alpha+\frac{\pi}{12}\right)=\frac{\sin \alpha \sin \left(\alpha+\frac{\pi}{6}\right)}{\cos \alpha \cos \left(\alpha+\frac{\pi}{6}\right)}
$$

Since $\operatorname{tg}^{2} \frac{\alpha}{2}=\frac{1-\cos \alpha}{1+\cos \alpha}, \sin \alpha \sin \beta=\frac{1}{2}(\cos (\alpha-\beta)-\cos (\alpha+\beta))$ and $\cos \alpha \cos \beta=\frac{1}{2}(\cos (\alpha-\beta)+\cos (\alpha+\beta))$, we have

$$
\begin{aligned}
& \frac{1-\cos \left(2 \alpha+\frac{\pi}{6}\right)}{1+\cos \left(2 \alpha+\frac{\pi}{6}\right)}=\frac{\cos \frac{\pi}{6}-\cos \left(2 \alpha+\frac{\pi}{6}\right)}{\cos \frac{\pi}{6}+\cos \left(2 \alpha+\frac{\pi}{6}\right)} \\
& \frac{1-\cos \left(2 \alpha+\frac{\pi}{6}\right)}{1+\cos \left(2 \alpha+\frac{\pi}{6}\right)}=\frac{\sqrt{3}-2 \cos \left(2 \alpha+\frac{\pi}{6}\right)}{\sqrt{3}+2 \cos \left(2 \alpha+\frac{\pi}{6}\right)} \Rightarrow \\
& \Rightarrow\left(1-\cos \left(2 \alpha+\frac{\pi}{6}\right)\right)\left(\sqrt{3}+2 \cos \left(2 \alpha+\frac{\pi}{6}\right)\right)=\left(\sqrt{3}-2 \cos \left(2 \alpha+\frac{\pi}{6}\right)\right) \times \\
& \times\left(1+\cos \left(2 \alpha+\frac{\pi}{6}\right)\right) \Leftrightarrow
\end{aligned}
$$

$\Leftrightarrow \cos \left(2 \alpha+\frac{\pi}{6}\right)=0,2 \alpha+\frac{\pi}{6}=\frac{\pi}{2}, 2 \alpha=\frac{\pi}{2}-\frac{\pi}{6}=\frac{2 \pi}{6}, \alpha=\frac{\pi}{6}$.

Then $\beta=\frac{\pi}{6}+\frac{\pi}{12}=\frac{\pi}{4}, \gamma=\frac{\pi}{6}+\frac{\pi}{6}=\frac{\pi}{3}$.

Answer: $\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}$.

Solve the systems of equations (8.394-8.405):