Task B-3.1. Solve the equation

$$
\log _{\sqrt{2} \sin x}(1+\cos x)=2
$$
Solution. The equation makes sense only if $1+\cos x>0$, which means $\cos x>-1$, and this is true for all $x \in \mathbb{R}$ except when $\cos x=-1$.

(1 point)

Similarly, the base of the logarithmic function must be positive and different from 1, so we have $\sqrt{2} \sin x>0$ and $\sqrt{2} \sin x \neq 1$, that is,

$$
\sin x>0 \quad \text { and } \quad \sin x \neq \frac{1}{\sqrt{2}}
$$

From the equation itself, we get

$$
\begin{aligned}
1+\cos x & =2 \sin ^{2} x \\
1+\cos x & =2\left(1-\cos ^{2} x\right) \\
2 \cos ^{2} x+\cos x-1 & =0
\end{aligned}
$$

thus we obtain $\cos x=\frac{1}{2}$ and $\cos x=-1$.

(1 point)

$\cos x=-1$ is not a solution due to the initial condition, and from $\cos x=\frac{1}{2}$ we get $x=\frac{\pi}{3}+2 k \pi$ or $x=-\frac{\pi}{3}+2 k \pi$. Due to the initial condition, the solution is only $x=\frac{\pi}{3}+2 k \pi$.

(1 point)