A triangle has sides $a, b, c$ and the angle opposite to $b$ is $\beta$. What can we say about the angles of the triangle if

$$
c^{2}=a^{2}+2 b^{2} \cos \beta
$$
Let $a$ be the side opposite angle $\alpha$, $c$ be the side opposite angle $\gamma$, and $r$ be the radius of the circumcircle of the triangle. Substituting the following relationships into the condition

$$
c^{2}=a^{2}+2 b^{2} \cos \beta
$$

we get

$$
a=2 r \sin \alpha, \quad b=2 r \sin \beta, \quad c=2 r \sin \gamma
$$

which yields the condition equivalent to (1):

$$
\sin ^{2} \gamma-\sin ^{2} \alpha=2 \sin ^{2} \beta \cos \beta
$$

Here,

$$
2 \sin \beta \cos \beta=\sin 2 \beta \quad \text { and } \sin ^{2} \gamma-\sin ^{2} \alpha= \\
=\left(\sin ^{2} \gamma-\sin ^{2} \gamma \sin ^{2} \alpha\right)-\left(\sin ^{2} \alpha-\sin ^{2} \alpha \sin ^{2} \gamma\right)= \\
=\sin ^{2} \gamma \cos ^{2} \alpha-\sin ^{2} \alpha \cos ^{2} \gamma=\sin (\gamma+\alpha) \sin (\gamma-\alpha)
$$

Since $\sin (\gamma+\alpha)=\sin \beta \neq 0$, we finally get the condition equivalent to (1) and (2):

$$
\sin (\gamma-\alpha)=\sin 2 \beta
$$

This generally means that

$$
\gamma-\alpha=2 \beta+k \cdot 360^{\circ}
$$

or

$$
\gamma-\alpha+2 \beta=180^{\circ}+k \cdot 360^{\circ}
$$

where $k$ is an appropriate integer. However, since in our case $\alpha, \beta, \gamma$ are the angles of a triangle, $k$ can only be 0 in both cases. From (4a), substituting $\alpha=180-\beta-\gamma$, we get

$$
\gamma=\frac{\beta}{2}+90^{\circ}, \quad \alpha=90^{\circ}-\frac{3}{2} \beta, \quad 0^{\circ}<\beta<60^{\circ}
$$

From (4b), substituting $\gamma=180^{\circ}-\beta-\alpha$, we get

$$
\alpha=\frac{\beta}{2}, \quad \gamma=180^{\circ}-\frac{3}{2} \beta, \quad 0^{\circ}<\beta<120^{\circ}
$$

Remark. This problem is related to problem 2057 in the "Comprehensive Collection of Mathematical Problems" workbook, which, due to a printing error, asks to prove that if $3 \beta+2 \gamma=180^{\circ}$, then $a^{2}-c^{2}=(1+c) b^{2}$, whereas $c$ should be replaced with $c / a$ on the right-hand side.