## Task B-3.2.

How many solutions does the equation $\frac{1}{\cos ^{2} \frac{x}{2}}+\frac{1}{\sin ^{2} \frac{x}{2}}=\frac{16}{\operatorname{tg} x}$ have on the interval $\langle 0,2019 \pi\rangle$?
## Solution.

The condition for the solution of the given equation is $\operatorname{tg} x \neq 0, \sin \frac{x}{2} \neq 0, \cos \frac{x}{2} \neq 0$. In other words, it must be $x \neq k \pi$ for $k \in \mathbb{Z}$.

Reduce the expression on the left side of the equation to a common denominator.

$$
\frac{1}{\cos ^{2} \frac{x}{2}}+\frac{1}{\sin ^{2} \frac{x}{2}}=\frac{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}}{\cos ^{2} \frac{x}{2} \cdot \sin ^{2} \frac{x}{2}}=\frac{1}{\cos ^{2} \frac{x}{2} \cdot \sin ^{2} \frac{x}{2}}
$$

Thus, the equation becomes

$$
\begin{gathered}
\frac{1}{\cos ^{2} \frac{x}{2} \cdot \sin ^{2} \frac{x}{2}}=\frac{16}{\operatorname{tg} x} \\
\frac{1}{\left(\frac{1}{2} \sin \left(2 \cdot \frac{x}{2}\right)\right)^{2}}=\frac{16}{\operatorname{tg} x} \\
\frac{1}{\frac{1}{4} \sin ^{2} x}=\frac{16 \cos x}{\sin x}
\end{gathered}
$$

Given the condition $\sin x \neq 0$, this is equivalent to $4 \sin x \cos x=1$, or $\sin 2 x=\frac{1}{2}$.

Solving this equation, we get $2 x=\frac{\pi}{6}+2 k \pi$ or $2 x=\frac{5 \pi}{6}+2 k \pi$, for $k \in \mathbb{Z}$, or $x=\frac{\pi}{12}+k \pi$ or $x=\frac{5 \pi}{12}+k \pi$, for $k \in \mathbb{Z}$.

The given equation has two solutions on each interval $\langle k \pi,(k+1) \pi\rangle$.

Since there are two solutions of the given equation on the interval $\langle 0, \pi\rangle$, there are a total of 4038 solutions on the interval $\langle 0,2019 \pi\rangle$.