8.126. $\frac{1}{1+\cos ^{2} z}+\frac{1}{1+\sin ^{2} z}=\frac{16}{11}$.
Solution.

Rewrite the given equation as

$$
\begin{aligned}
& \frac{1}{1+\frac{1}{2}(1+\cos 2 z)}+\frac{1}{1+\frac{1}{2}(1-\cos 2 z)}-\frac{16}{11}=0, \frac{2}{3+\cos 2 z}+\frac{2}{3-\cos 2 z}-\frac{16}{11}=0 \Rightarrow \\
& \Rightarrow \cos ^{2} 2 z=\frac{3}{4}, \cos 2 z= \pm \frac{\sqrt{3}}{2}, 2 z= \pm \frac{\pi}{6}+\pi k \\
& z= \pm \frac{\pi}{12}+\frac{\pi k}{2}=\frac{\pi}{12}(6 k \pm 1), k \in Z
\end{aligned}
$$

Answer: $\quad z=\frac{\pi}{12}(6 k \pm 1), \quad k \in Z$.

$$
\text { 8.127. } \frac{\cos x}{2 \operatorname{tg} \frac{x}{2} \cos ^{2} \frac{x}{2}}-\frac{\operatorname{tg} \frac{x}{2} \operatorname{tg} x+1}{\operatorname{tg} \frac{x}{2}+\operatorname{ctg} x}=2 \sqrt{3}
$$

## Solution.

$$
\text { Domain: }\left\{\begin{array}{l}
\cos \frac{x}{2} \neq 0 \\
\sin x \neq 0 \\
\operatorname{tg} \frac{x}{2}+\operatorname{ctg} x \neq 0
\end{array}\right.
$$

From the condition, we get

$$
\frac{\cos x}{\frac{2 \sin \frac{x}{2} \cos ^{2} \frac{x}{2}}{\cos \frac{x}{2}}}-\frac{\frac{\sin \frac{x}{2} \sin x}{\cos \frac{x}{2} \cos x}}{\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}+\frac{\cos x}{\sin x}}=2 \sqrt{3} \Leftrightarrow
$$

$\Leftrightarrow \frac{\cos x}{\sin x}-\frac{\sin \frac{x}{2} \sin x+\cos \frac{x}{2} \cos x}{\cos \frac{x}{2} \cos x} \cdot \frac{\cos \frac{x}{2} \sin x}{\sin \frac{x}{2} \sin x+\cos \frac{x}{2} \cos x}=2 \sqrt{3} \Leftrightarrow$
$\Leftrightarrow \frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}=2 \sqrt{3} \Leftrightarrow \frac{\cos ^{2} x-\sin ^{2} x}{\sin x \cos x}=2 \sqrt{3}$,

$\frac{\cos ^{2} x-\sin ^{2} x}{2 \sin x \cos x}=\sqrt{3}, \operatorname{ctg} 2 x=\sqrt{3}, 2 x=\frac{\pi}{6}+\pi k$,

$x=\frac{\pi}{12}+\frac{\pi k}{2}=\frac{\pi}{12}(6 k+1), k \in Z$.

Answer: $x=\frac{\pi}{12}(6 k+1), k \in Z$.