Solve the equation

$$
\cos x=\frac{\sin x}{1-\sin x}
$$
Given that $1-\sin x \neq 0$, our equation can also be written as:

$$
(1-\sin x) \cos x=\sin x
$$

which, when rearranged, gives

$$
\cos x-\sin x=\sin x \cdot \cos x
$$

Squaring both sides, we get

$$
\cos ^{2} x-2 \sin x \cos x+\sin ^{2} x=\sin ^{2} x \cdot \cos ^{2} x
$$

Using the identities $\sin ^{2} x+\cos ^{2} x=1$ and $2 \sin x \cos x=\sin 2 x$, we obtain

$$
1-\sin 2 x=\frac{\sin ^{2} 2 x}{4}
$$

Rearranging, we have

$$
\sin ^{2} 2 x+4 \sin 2 x-4=0
$$

from which

$$
\sin 2 x=-2 \pm 2 \sqrt{2}
$$

Since $|\sin 2 x| \leq 1$, only the positive sign provides a solution, so

$$
\sin 2 x=2 \sqrt{2}-2 \sim 0.8284
$$

From this,

$$
\begin{array}{ll}
2 x_{1}=55^{\circ} 56^{\prime}+k \cdot 360^{\circ}, & x_{1}=27^{\circ} 58^{\prime}+k \cdot 180 \\
2 x_{2}=124^{\circ} 4^{\prime}+k \cdot 360^{\circ}, & x_{2}=62^{\circ} 2^{\prime}+k \cdot 180
\end{array}
$$

When squaring, false roots may have been introduced. Let's exclude these. From the original equation, it is clear that $\sin x$ and $\cos x$ have the same sign because $1-\sin x$ is always positive. Therefore, $\sin x \cdot \cos x$ is always positive, and thus (1) implies that

$$
\begin{aligned}
& \cos x-\sin x>0 \\
& \cos 27^{\circ} 58^{\prime}-\sin 27^{\circ} 58^{\prime}>0 \\
& \cos \left(180^{\circ}+27^{\circ} 58^{\prime}\right)-\sin \left(180^{\circ}+27^{\circ} 58^{\prime}\right)>0
\end{aligned}
$$

Thus, only

$$
x_{1}=27^{\circ} 58^{\prime}+k \cdot 360
$$

and

$$
x_{2}=242^{\circ} 2^{\prime}+k \cdot 360
$$

can be solutions to the equation. Substitution confirms that both values satisfy the equation.

Csiszár Imre (Bp. I., Petőfi g. IV. o. t.)