1. Compute $\operatorname{tg} \alpha$, if $3 \operatorname{tg} \alpha-\sin \alpha+4 \cos \alpha=12$.
4.II.1. We have,

$$
\begin{aligned}
& 3 \operatorname{tg} \alpha - \sin \alpha + 4 \cos \alpha = 12 \\
& \Longleftrightarrow \operatorname{tg} \alpha (3 - \cos \alpha) - 4 (3 - \cos \alpha) = 0 \Longleftrightarrow \\
& (\operatorname{tg} \alpha - 4)(3 - \cos \alpha) = 0 \Longleftrightarrow \operatorname{tg} \alpha = 4
\end{aligned}
$$

Answer: 4.