## Task $6 / 72$

In the equation

$$
\cot (m \cos (2 \pi x))=\sqrt{3}
$$

the coefficient $m>0$ is to be determined such that the equation has the solutions $x_{1 ; 2}= \pm \frac{1}{6}$. Furthermore, all other solutions of the equation are to be determined.
If we set $x_{1 ; 2}= \pm \frac{1}{6}$ in the given equation, we obtain

$$
\cot \left(m \cos \frac{\pi}{3}\right)=\sqrt{3}
$$

since $\cos (-x)=\cos x$. From this, we further get $\cos \frac{\pi}{3}=\frac{1}{2} \rightarrow \cot \frac{m}{2}=\sqrt{3}$, hence $m=\frac{\pi}{3}$.

Equation (1) then takes the form

$$
\cot \left(\frac{\pi}{3} \cos (2 \pi x)\right)=\sqrt{3}
$$

From this, it follows that

$$
\frac{\pi}{3} \cos (2 \pi x)=\operatorname{arccot}(\sqrt{3})=\frac{\pi}{6}+k \pi
$$

with $k=0 ; \pm 1 ; \pm 2 ; \ldots$ Multiplying by $\frac{3}{\pi}$, we get $\cos (2 \pi x)=3 k+\frac{1}{2}$.

Since $|\cos (2 \pi x)| \leq 1$, $k$ must be 0, i.e., $\cos (2 \pi x)=\frac{1}{2}$, thus $2 \pi x=\arccos \frac{1}{2}= \pm \frac{\pi}{3}+2 n \pi$ with $n=0 ; \pm 1 ; \pm 2 ; \ldots$ Therefore, we finally get $x= \pm \frac{1}{6}+n$.