
3. Find the magnitudes of the interior angles $\alpha, \beta, \gamma$ of a triangle which satisfy

$$
\begin{aligned}
& 2 \sin \beta \sin (\alpha+\beta)-\cos \alpha=1 \\
& 2 \sin \gamma \sin (\beta+\gamma)-\cos \beta=0
\end{aligned}
$$


Solution. Similarly as in the solution of the first problem of the First Round we can derive the following formula

$$
\cos x-2 \sin y \sin (x+y)=\cos (x+2 y)
$$

which holds for arbitrary real numbers $x, y$. From conditions of the given problem we get the following two equations for magnitudes of the interior angles $\alpha, \beta, \gamma$ of a triangle:

$$
\begin{aligned}
& \cos \alpha-2 \sin \beta \sin (\alpha+\beta)=\cos (\alpha+2 \beta)=-1, \\
& \cos \beta-2 \sin \gamma \sin (\beta+\gamma)=\cos (\beta+2 \gamma)=0 .
\end{aligned}
$$

This yields

$$
\begin{aligned}
& \alpha+2 \beta=\pi+2 k \pi, \\
& \beta+2 \gamma=\frac{\pi}{2}+l \pi,
\end{aligned}
$$

where $k, l$ are arbitrary non-negative integers. Adding up (1) and (2) we obtain

$$
\alpha+3 \beta+2 \gamma=\frac{3 \pi}{2}+(2 k+l) \pi
$$

Since $\alpha, \beta, \gamma$ are magnitudes of interior angles of a triangle $(\alpha+\beta+\gamma=\pi)$ we have

$$
2 \beta+\gamma=\frac{\pi}{2}+(2 k+l) \pi
$$

where $k, l$ are non-negative integers. Further from (2) a (3) we obtain after short manipulation

$$
\beta+\gamma=\frac{\pi}{3}+\frac{2}{3}(k+l) \pi
$$

Regarding $\beta+\gamma0$. Thus $\sin \beta=\frac{1}{2}$ (the possibility $\sin \beta=-\frac{1}{2}$ is ruled out for an angle $\beta$ in the interval $(0, \pi)$ ). We have thus arrived at the only possible values $\beta=\gamma=30^{\circ}$, implying $\alpha=120^{\circ}$. These values are indeed a solution of the original system, since the first equation holds owing to the equality $\beta=\gamma$, while the second has been treated under the assumption of $\beta=\gamma$ only by equivalent manipulations.
