## Task A-3.1.

Find all solutions to the equation

$$
\operatorname{ctg}\left(2 \pi \cos ^{2}(2 \pi x)\right)=0
$$
## First Solution.

The equality $\operatorname{ctg}\left(2 \pi \cos ^{2}(2 \pi x)\right)=0$ is satisfied if and only if

$$
2 \pi \cos ^{2}(2 \pi x)=\frac{\pi}{2}+k \pi, \quad \text { for some } k \in \mathbb{Z}
$$

or equivalently

$$
\cos ^{2}(2 \pi x)=\frac{1}{4}+\frac{k}{2}, \quad k \in \mathbb{Z}
$$

The values of the function $\cos ^{2}(2 \pi x)$ are in the interval $[0,1]$, so $k$ can only be 0 or 1.

For $k=0$, we get the equation $\cos ^{2}(2 \pi x)=\frac{1}{4}$,

so $\cos (2 \pi x)=\frac{1}{2}$ or $\cos (2 \pi x)=-\frac{1}{2}$.

The solutions to the equation $\cos (2 \pi x)=\frac{1}{2}$ are:

$$
x \in\left\{\left.\frac{1}{6}+m \right\rvert\, m \in \mathbb{Z}\right\} \cup\left\{\left.\frac{5}{6}+m \right\rvert\, m \in \mathbb{Z}\right\}
$$

and the solutions to the equation $\cos (2 \pi x)=-\frac{1}{2}$ are:

$$
x \in\left\{\left.\frac{1}{3}+m \right\rvert\, m \in \mathbb{Z}\right\} \cup\left\{\left.\frac{2}{3}+m \right\rvert\, m \in \mathbb{Z}\right\}
$$

The solutions found so far can also be written as:

$$
x \in\left\{\left.\frac{1}{6}+\frac{m}{2} \right\rvert\, m \in \mathbb{Z}\right\} \cup\left\{\left.\frac{1}{3}+\frac{m}{2} \right\rvert\, m \in \mathbb{Z}\right\}
$$

For $k=1$, we get the equation $\cos ^{2}(2 \pi x)=\frac{3}{4}$,

so $\cos (2 \pi x)=\frac{\sqrt{3}}{2}$ or $\cos (2 \pi x)=-\frac{\sqrt{3}}{2}$.

The solutions to the equation $\cos (2 \pi x)=\frac{\sqrt{3}}{2}$ are:

$$
x \in\left\{\left.\frac{1}{12}+m \right\rvert\, m \in \mathbb{Z}\right\} \cup\left\{\left.\frac{11}{12}+m \right\rvert\, m \in \mathbb{Z}\right\}
$$

and the solutions to the equation $\cos (2 \pi x)=-\frac{\sqrt{3}}{2}$ are:

$$
x \in\left\{\left.\frac{5}{12}+m \right\rvert\, m \in \mathbb{Z}\right\} \cup\left\{\left.\frac{7}{12}+m \right\rvert\, m \in \mathbb{Z}\right\}
$$

These solutions can also be written as:

$$
x \in\left\{\left.\frac{1}{12}+\frac{m}{2} \right\rvert\, m \in \mathbb{Z}\right\} \cup\left\{\left.\frac{5}{12}+\frac{m}{2} \right\rvert\, m \in \mathbb{Z}\right\}
$$

The set of all solutions to the given equation can be written more simply as:

$$
x \in\left\{\left.\frac{1}{12}+\frac{m}{4} \right\rvert\, m \in \mathbb{Z}\right\} \cup\left\{\left.\frac{1}{6}+\frac{m}{4} \right\rvert\, m \in \mathbb{Z}\right\}
$$