Find the value of $\operatorname{ctg} x$ if $\operatorname{ctg} x = \sin x?$
Solution. We know that $1+\operatorname{ctg}^{2} x=\frac{1}{\sin ^{2} x}$. Since ctg $x=\sin x$, then

$$
1+\operatorname{ctg}^{2} x=\frac{1}{\operatorname{ctg}^{2} x}
$$

Rearranging: $\operatorname{ctg}^{4} x+\operatorname{ctg}^{2} x-1=0$, from which

$$
\operatorname{ctg}^{2} x=\frac{\sqrt{5}-1}{2}, \quad \text { or } \quad \operatorname{ctg}^{2} x=\frac{-\sqrt{5}-1}{2}
$$

The latter cannot be true, since the square of any number is non-negative.

Thus $\operatorname{ctg}^{2} x=\frac{\sqrt{5}-1}{2}$, therefore

$$
\operatorname{ctg} x=\sqrt{\frac{\sqrt{5}-1}{2}}, \quad \text { or } \quad \operatorname{ctg} x=-\sqrt{\frac{\sqrt{5}-1}{2}}
$$

Of course, these transformations only make sense if $\sin x \neq 0$, i.e., $x \neq k \pi(k \in \mathbb{Z})$, and $\operatorname{ctg} x \neq 0$, i.e., $x \neq \frac{\pi}{2}+n \pi(n \in \mathbb{Z})$. It is easily seen that these "constraints" do not narrow the set of solutions.