3. Determine the sizes of the internal angles $\alpha, \beta, \gamma$ of a triangle for which the following conditions hold:

$$
\begin{aligned}
& 2 \sin \beta \sin (\alpha+\beta)-\cos \alpha=1 \\
& 2 \sin \gamma \sin (\beta+\gamma)-\cos \beta=0
\end{aligned}
$$
3. From the equality $\alpha+\beta+\gamma=\pi$ and known trigonometric formulas, we get

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

Substitute these expressions for $\sin (\alpha+\beta)$ and $\cos \alpha$ into the first equation from the problem and simplify:

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

The last equality holds if and only if $\beta=\gamma$, because the difference of two interior angles of a triangle lies in the interval $(-\pi, \pi)$, in which the cosine function takes the value 1 only at the point zero. Thus, we have shown that the first given equation is satisfied for the interior angles of a triangle if and only if $\beta=\gamma$.

Now we can easily solve the second given equation by substituting $\gamma$ with $\beta$:

$$
\begin{aligned}
2 \sin \beta \sin 2 \beta-\cos \beta & =0, \\
4 \sin ^{2} \beta \cos \beta-\cos \beta & =0, \\
\left(4 \sin ^{2} \beta-1\right) \cos \beta & =0 .
\end{aligned}
$$

Thus, either $\cos \beta=0$ or $\sin \beta= \pm \frac{1}{2}$. The equality $\beta=\gamma$ for the angles of a triangle means that the angle $\beta$ is acute, so $\cos \beta>0$, and therefore $\sin \beta=\frac{1}{2}$ (the value $\sin \beta=-\frac{1}{2}$ is excluded for $\beta$ in the interval $(0, \pi)$). Thus, we arrive at the only possible values $\beta=\gamma=30^{\circ}$, from which we easily find $\alpha=120^{\circ}$. In this process, verification is not necessary: the first given equation holds due to the equality $\beta=\gamma$, and the second equation was solved using equivalent transformations under the assumption $\beta=\gamma$.

Another solution. Similarly to the solution of the home round problem, we use known trigonometric formulas to derive the equality

$$
\begin{aligned}
2 \sin y \sin (x+y)-\cos x & =2 \sin y(\sin x \cos y+\cos x \sin y)-\cos x= \\
& =2 \sin y \cos y \sin x+\left(2 \sin ^{2} y-1\right) \cos x= \\
& =\sin 2 y \sin x-\cos 2 y \cos x= \\
& =-\cos (x+2 y)
\end{aligned}
$$

for any real numbers $x, y$. This allows us to rewrite the system of equations from the problem as

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

The interior angles of any triangle lie in the interval $(0, \pi)$, from which the inequalities $0<\alpha+2 \beta<3 \pi$ follow. From these, it follows that the equation (1) is satisfied if and only if $\alpha+2 \beta=\pi$. Comparing this with the general equality $\alpha+\beta+\gamma=\pi$, we get the equivalent condition $\gamma=\beta$, under which (2) becomes

$$
\cos 3 \beta=0 \text {. }
$$

Since the angle $\beta$ is acute (as it is equal to $\gamma$ and a triangle cannot have two right or two obtuse interior angles), the inequalities $0<3 \beta<\frac{3}{2} \pi$ hold, under which the equation (3) is satisfied if and only if $3 \beta=\frac{1}{2} \pi$ or $\beta=\gamma=\frac{1}{6} \pi$. As in the first solution, we find $\alpha=\pi-\beta-\gamma=\frac{2}{3} \pi$. Verification (which is not necessary in this process) easily confirms that the found triplet of angles $\alpha, \beta, \gamma$ satisfies all conditions of the problem.

The conditions of the problem are satisfied only by triangles whose interior angles have the measures $\alpha=120^{\circ}, \beta=\gamma=30^{\circ}$.

Award 6 points for a complete solution. For the first approach, award 3 points for deriving the relationship $\beta=\gamma$ from the first equation, 1 point for transforming the second equation into a product form or a basic trigonometric equation (such as $\cos 3 \beta=0$), and the remaining 2 points for subsequently determining (with justification for uniqueness) the measures of all three angles.

For the second approach, award 2 points for deriving the system of equations (1) and (2) (in their basic form), another 2 points for justifying the relationship $\beta=\gamma$, and the remaining 2 points as in the first approach.

If the solver's approach requires verification, deduct 1 point for its absence. Also deduct 1 point each time any of the values $\beta-\gamma, \alpha+2 \beta, 3 \beta$ etc. are determined from the corresponding basic trigonometric equation without mentioning the necessary inequalities for the represented argument.