Task B-3.5. For which positive real numbers $a$ does the equation

$$
\frac{1}{\sin ^{2} x}+\frac{1}{\cos ^{2} x}+\frac{1}{\operatorname{tg}^{2} x}+\frac{1}{\operatorname{ctg}^{2} x}=a, \quad x \neq \frac{k \pi}{2}, \quad k \in \mathbb{Z}
$$

have real solutions?
## Solution.

Let's rewrite the given equation in another form:

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

from which it follows that

$$
\sin ^{2} x \cos ^{2} x=\frac{2}{2+a}, \quad \sin ^{2} 2 x=\frac{8}{2+a}
$$

This equation will have real solutions if

$$
0 \leq \frac{8}{2+a} \leq 1
$$

Since $a>0$, this inequality is true for all $a \geq 6$.