# 2. Solution:

$\sin ^{4} 2019 x+\cos ^{2019} 2022 x \cdot \cos ^{2018} 2019 x=1 \Leftrightarrow\left(1-\cos ^{2} 2019 x\right)^{2}+\cos ^{2019} 2022 x \cdot \cos ^{2018} 2019 x=1 \Leftrightarrow$ $\cos ^{4} 2019 x-2 \cos ^{2} 2019 x+\cos ^{2019} 2022 x \cdot \cos ^{2018} 2019 x=0 \Leftrightarrow$ $\cos ^{2} 2019 x\left(\cos ^{2} 2019 x+\cos ^{2019} 2022 x \cdot \cos ^{2016} 2019 x-2\right)=0 \Leftrightarrow$ 1) $\cos 2019 x=0, x=\frac{\pi}{4038}+\frac{\pi n}{2019}, n \in Z$ or 2) $\cos ^{2} 2019 x+\cos ^{2019} 2022 x \cdot \cos ^{2016} 2019 x=2 \Leftrightarrow$ $\left\{\begin{array}{c}\cos ^{2} 2019 x=1, \\ \cos ^{2019} 2022 x \cdot \cos ^{2016} 2019 x=1,\end{array} \Leftrightarrow\left\{\begin{array}{l}\sin 2019 x=0, \\ \cos 2022 x=1,\end{array} \Leftrightarrow\left\{\begin{array}{l}x=\frac{\pi n}{2019}, n \in Z, \\ x=\frac{\pi m}{1011}, m \in Z,\end{array}\right.\right.\right.$ $\frac{n}{2019}=\frac{m}{1011}, n=\frac{2019 m}{1011}=\frac{673 m}{337}, n \in Z, m \in Z \Rightarrow m=337 k, \quad x=\frac{\pi k}{3}, k \in Z$
Answer: $x=\frac{\pi}{4038}+\frac{\pi n}{2019}, n \in Z, x=\frac{\pi k}{3}, k \in Z$.

![](https://cdn.mathpix.com/cropped/2024_05_06_e1272d04810341416431g-19.jpg?height=66&width=1659&top_left_y=1212&top_left_x=313)
$O_{2}$ is the center of the inscribed circle in triangle $A D E$, and $r$ is its radius. Let $p_{1}$ be the semiperimeter of triangle $A B C$, and $p_{2}$ be the semiperimeter of triangle $A D E$. Then $A K=p_{2}, A K=p_{1}-B C$. Indeed, by the properties of tangents to a circle, we have (see the figure)

$$
\begin{gathered}
2 A K=A K+A H=A D+D K+E H+A E=A D+(D M+M E)+A E=2 p_{2} \\
2 A K=(A B-B K)+(A C-C H)=A B+A C+(B F+C F)-2(B F+C F)=2\left(p_{1}-B C\right) .
\end{gathered}
$$

Then $r=S_{A D E} / p_{2}=S_{A D E} / A K=1 / 18$.

Since a circle can be circumscribed around quadrilateral $B D E C$, we have $\angle A B C+\angle D E C=180^{\circ}$, and $\angle A E D=\angle A B C$.

Therefore, triangles $A D E$ and $A C B$ are similar, and $\frac{r}{R}=\frac{p_{2}}{p_{1}}$, $R=\frac{r(A K+B C)}{A K}=\frac{1+5}{18}=\frac{1}{3}$.

Let $\angle B A C=\alpha$. Then

$\operatorname{tg}(\alpha / 2)=R / A K=1 / 3$,

$\operatorname{tg} \alpha=2 \operatorname{tg}(\alpha / 2) /\left(1-\operatorname{tg}^{2}(\alpha / 2)\right)=3 / 4$.

![](https://cdn.mathpix.com/cropped/2024_05_06_e1272d04810341416431g-19.jpg?height=617&width=880&top_left_y=1879&top_left_x=1065)

Answer: $3 / 4$.