Task A-3.2. (8 points)

Solve the inequality

$$
\log _{2}(1-2 \cos x)+\log _{\frac{1}{2}}(1+2 \cos x) \leq 0
$$

in the interval $[0,2 \pi]$.
## Solution.

For the logarithms to be defined, it must hold that: $1-2 \cos x>0$ and $1+2 \cos x>0$, which means $-\frac{1}{2}<\cos x<\frac{1}{2}$.

Using the properties of logarithms, the equation can be written in the form: $\log _{2} \frac{1-2 \cos x}{1+2 \cos x} \leq 0$, or $\frac{1-2 \cos x}{1+2 \cos x} \leq 1$.

The denominator of this fraction is positive (due to the condition), so after rearranging, the inequality becomes $\cos x \geq 0$.

The solution to the given inequality in the interval $[0,2 \pi]$ is:

$$
x \in\left\langle\frac{\pi}{3}, \frac{\pi}{2}\right] \cup\left[\frac{3 \pi}{2}, \frac{5 \pi}{3}\right\rangle
$$

For 4 points: conditions and removing the logarithm (*) or solved to the end while neglecting the conditions (result: $x \in\left[0, \frac{\pi}{2}\right] \cup\left[\frac{3 \pi}{2}, 2 \pi\right]$ ).

For 8 points: correct result and procedure.