## Task B-4.2.

Determine all real numbers $x$ for which the numbers $\log (\sin x), \log (\sin x+1)$, and $\log (3 \sin x+1)$ are, in that order, consecutive terms of an arithmetic sequence.
## Solution.

Since the logarithm is defined only for positive real numbers, the condition for the problem to have a solution is $\sin x>0$.

The given numbers are consecutive terms of an arithmetic sequence, so the sought numbers $x$ are solutions to the equation:

$$
\log (3 \sin x+1)+\log (\sin x)=2 \log (\sin x+1)
$$

Then we have:

$$
\log [(3 \sin x+1) \sin x]=\log (\sin x+1)^{2}
$$

or

$$
(3 \sin x+1) \sin x=(\sin x+1)^{2}
$$

This leads to a quadratic equation in $\sin x$:

$$
2 \sin ^{2} x-\sin x-1=0
$$

Then $\sin x=-0.5$ or $\sin x=1$.

We discard the solution $\sin x=-0.5$ due to the conditions of the problem, and from the solution $\sin x=1$ it follows that the sought real numbers are of the form:

$$
x=\frac{\pi}{2}+2 k \pi, k \in \mathbf{Z}
$$

Note: Points should not be deducted if the student did not state the condition but discarded the possibility that $\sin x=-0.5$ and obtained the correct solution.