To solve the following equation:

$$
\frac{1}{\sin x}-\cot x-\sin x=0
$$

Does the curve of the function $y=\frac{1}{\sin x}-\cot x-\sin x$ pass through the origin?

Calculate the value of $y'$ when $x=0$ and $x=\frac{\pi}{2}$.
$1^{0}$. For the time being, let's assume that $\sin x \neq 0$. We can then write our equation as

$$
\frac{1}{\sin x}-\frac{\cos x}{\sin x}-\sin x=0, \text { or } \frac{1-\cos x-\sin ^{2} x}{\sin x}=0
$$

This equation is satisfied by the roots of

$$
1-\cos x-\sin ^{2} x=0 \text { or } 1-\cos x-\left(1-\cos ^{2} x\right)=\cos x(\cos x-1)=0
$$

Thus,

$$
\cos x=0 \text { or } x=(2 k+1) \frac{\pi}{2} \text { and } \cos x=1, \text { or } x=2 k \pi
$$

However, if $x=2 k \pi$, then $\sin x=0$. We will examine this solution, which we have temporarily excluded, separately.

Our equation can also be written as:

$$
\frac{1-\cos x}{\sin x}-\sin x=\frac{2 \sin ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}-\sin x=2 \operatorname{tg} \frac{x}{2}-\sin x=0
$$

If $\sin x=0$ already, then $\operatorname{tg} \frac{x}{2}=0$.

From this,

$$
\frac{x}{2}=k \pi, \quad x=2 k \pi
$$

Therefore, the solutions to the given equation are: $x=(2 k+1) \frac{\pi}{2}$, $x=2 k \pi$, where $k$ is any integer, or zero.

$2^{0}$. The curve of $y$ intersects the $X$-axis at the points where $y=0$, i.e., at $x=2 k \pi$. If $k=0, x=0, y=0$, so the curve passes through the origin.

$3^{0} \cdot$ The derivative of $y=\frac{1-\cos x}{\sin x}-\sin x$ is

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

At $x=0, y^{\prime}=\frac{1}{2}-1=-\frac{1}{2}$.

At $x=\frac{\pi}{2}$, $\cos x=0$ and $\cos \frac{x}{2}=\cos \frac{\pi}{4}=\frac{\sqrt{2}}{2}$, so $2 \cos ^{2} \frac{x}{2}=1$ and thus

$$
y^{\prime}=1
$$

Note. A significant part of the solutions did not pay sufficient attention to the fact that at $x=0$,

$$
y=\frac{1}{\sin 0}-\operatorname{cotg} 0-\sin 0=\infty-\infty
$$

which is indeterminate. If we pay attention to the fact that $\frac{1}{\sin x}=\operatorname{cosec} x$ and examine the variation of trigonometric functions in relation to the unit circle, we will see that as $x$ approaches zero, the values of $\operatorname{cosec} x$ and $\operatorname{cotg} x$ (hypotenuse and adjacent side) approach each other and become equal near $x=0$. Therefore, there is no need for more advanced tools, such as L'Hôpital's rule or series expansion.