Solve the following equations:

$$
\operatorname{tg} x + \sin x = 1
$$

and

$$
\operatorname{tg} x - \sin x = 1
$$
(1)-from

$$
\begin{gathered}
\sin x+\sin x \cos x=\cos x \\
\frac{1}{2} \sin 2 x=\cos x-\sin x \\
\frac{1}{4} \sin ^{2} 2 x=(\cos x-\sin x)^{2}=1-2 \sin x \cos x \\
\sin ^{2} 2 x+4 \sin 2 x=4 \\
\sin 2 x=2(\sqrt{2}-1)
\end{gathered}
$$

We get the same from equation (2).

(3)-from:

$$
\begin{aligned}
& 2 x_{1}=55^{\circ} 56^{\prime} 20^{\prime \prime}, x_{1}=27^{\circ} 58^{\prime} 10^{\prime \prime} \\
& 2 x_{2}=124^{\circ} 3^{\prime} 40^{\prime \prime}, x_{2}=62^{\circ} 1^{\prime} 50^{\prime \prime}
\end{aligned}
$$

If we only consider acute angles, then from (l) it follows that

$$
\operatorname{tg} x>1$, so $x>45^{\circ}$,

therefore

$$
x_{2}=62^{\circ} 1^{\prime} 50^{\prime \prime}
$$

(Tóbiás J. László, Szeged.)

Number of solutions: 32.