8.1. For all triples $(x, y, z)$ satisfying the system

$$
\left\{\begin{array}{l}
2 \sin x=\operatorname{tg} y \\
2 \cos y=\operatorname{ctg} z \\
\sin z=\operatorname{tg} x
\end{array}\right.
$$

find the smallest value of the expression $\cos x-\sin z$.
Answer: $-\frac{5 \sqrt{3}}{6} \approx-1.44$.

Solution. Squaring the first equation and adding 1 to both sides, taking into account the second equation, we have

$$
4 \sin ^{2} x+1=\operatorname{tg}^{2} y+1=\frac{1}{\cos ^{2} y}=4 \operatorname{tg}^{2} z
$$

Performing the same procedure with the third equation, we find

$$
\sin ^{2} z+1=\operatorname{tg}^{2} x+1=\frac{1}{\cos ^{2} x} \quad \Rightarrow \quad 4 \cos ^{2} x=\frac{4}{\sin ^{2} z+1}
$$

Adding the obtained relations, we get

$$
5=4 \operatorname{tg}^{2} z+\frac{4}{\sin ^{2} z+1}=\frac{4 \sin ^{2} z}{1-\sin ^{2} z}+\frac{4}{\sin ^{2} z+1}=\frac{4+4 \sin ^{4} z}{1-\sin ^{4} z} \quad \Rightarrow \quad 9 \sin ^{4} z=1, \sin ^{2} z=\frac{1}{3}
$$

Substituting this value into the previous formulas, we calculate $\cos ^{2} x=\frac{3}{4}, \operatorname{tg}^{2} y=1$. Therefore, the solutions $(x, y, z)$ of the given system can only include the numbers

$$
x= \pm \frac{\pi}{6}+\pi k, \quad y= \pm \frac{\pi}{4}+\pi l, \quad z= \pm \arcsin \frac{1}{\sqrt{3}}+\pi m, k, l, m \in \mathbb{Z}
$$

and, therefore, only the following cases are possible

$$
\cos x= \pm \frac{\sqrt{3}}{2}, \quad \sin z= \pm \frac{1}{\sqrt{3}}
$$

Now let's analyze the system by the signs of the left and right parts:

- if $x$ lies in the I quadrant, then either $y$ lies in the I quadrant, $z$ lies in the I quadrant, or $y$ lies in the III quadrant, $z$ lies in the II quadrant;
- if $x$ lies in the II quadrant, then either $y$ lies in the I quadrant, $z$ lies in the III quadrant, or $y$ lies in the III quadrant, $z$ lies in the IV quadrant;
- if $x$ lies in the III quadrant, then either $y$ lies in the II quadrant, $z$ lies in the II quadrant, or $y$ lies in the IV quadrant, $z$ lies in the I quadrant;
- if $x$ lies in the IV quadrant, then either $y$ lies in the II quadrant, $z$ lies in the IV quadrant, or $y$ lies in the IV quadrant, $z$ lies in the III quadrant.

Taking the triplet $x=\frac{7 \pi}{6}, y=-\frac{\pi}{4}, z=\arcsin \frac{1}{\sqrt{3}}$ (by direct verification, we confirm that it satisfies the system from the problem statement), where $x$ is in the III quadrant and $z$ is in the I quadrant, we obtain the minimum of the possible values of $\cos x-\sin z$, which is $-\sqrt{3} / 2-1 / \sqrt{3}=-5 / 2 \sqrt{3}$.