## Task B-3.6.

Determine all real numbers $x$ in the interval $[0,1]$ for which $\operatorname{tg}\left(2 \pi \sin ^{2}(2 \pi x)\right)=0$.
## Solution.

The equality $\operatorname{tg}\left(2 \pi \sin ^{2} 2 \pi x\right)=0$ holds if $2 \pi \sin ^{2} 2 \pi x=k \pi, k \in \mathbb{Z}$ 1 point, or

$$
\sin ^{2} 2 \pi x=\frac{k}{2}
$$

due to the values of the sine function, $k$ can be $k=0, k=1, k=2$.

If $k=0$, then $\sin ^{2} 2 \pi x=0$, or $\sin 2 \pi x=0$, so we get

$$
2 \pi x=l \pi, l \in \mathbb{Z}
$$

It follows that $x_{1}=\frac{l}{2}, l \in \mathbb{Z}$.

If $k=1$, then $\sin ^{2} 2 \pi x=\frac{1}{2}$, or $\sin 2 \pi x= \pm \frac{\sqrt{2}}{2}$, so we get

$$
2 \pi x=\frac{\pi}{4}+\frac{m \pi}{2}, m \in \mathbb{Z}
$$

It follows that $x_{2}=\frac{1}{8}+\frac{m}{4}, m \in \mathbb{Z}$.

If $k=2$, then $\sin ^{2} 2 \pi x=1$, or $\sin 2 \pi x= \pm 1$, so we get

$$
2 \pi x=\frac{\pi}{2}+\frac{n \pi}{2}, n \in \mathbb{Z}
$$

It follows that $x_{3}=\frac{1}{4}+\frac{n}{4}, n \in \mathbb{Z}$.

Determine the required numbers in the interval $[0,1]$.

From $x_{1}$ we get the numbers $0, \frac{1}{2}, 1$, from $x_{2}$ the numbers $\frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}$, and from $x_{3}$ the numbers $\frac{1}{4}, \frac{3}{4}$.

Finally, $x \in\left\{0, \frac{1}{8}, \frac{1}{4}, \frac{3}{8}, \frac{1}{2}, \frac{5}{8}, \frac{3}{4}, \frac{7}{8}, 1\right\}$.