Решите уравнение |2tg x - 3| - |5tg x - 7| = -4.

Let $$y = \operatorname{tg} x$$. The equation becomes $$|2y - 3| - |5y - 7| = -4$$. We consider cases based on the signs of the expressions inside the absolute values.
Case 1: $$2y - 3 \ge 0$$ and $$5y - 7 \ge 0$$, which means $$y \ge 7/5$$. Then $$(2y - 3) - (5y - 7) = -4$$, so $$-3y + 4 = -4$$, which gives $$-3y = -8$$, so $$y = 8/3$$. Since $$8/3 \ge 7/5$$, this is a valid solution for $$y$$.
Case 2: $$2y - 3 < 0$$ and $$5y - 7 < 0$$, which means $$y < 3/2$$ and $$y < 7/5$$. So $$y < 7/5$$. Then $$-(2y - 3) - (-(5y - 7)) = -4$$, so $$-2y + 3 + 5y - 7 = -4$$, which gives $$3y - 4 = -4$$, so $$3y = 0$$, and $$y = 0$$. Since $$0 < 7/5$$, this is a valid solution for $$y$$.
Case 3: $$2y - 3 \ge 0$$ and $$5y - 7 < 0$$, which means $$3/2 \le y < 7/5$$. Then $$(2y - 3) - (-(5y - 7)) = -4$$, so $$2y - 3 + 5y - 7 = -4$$, which gives $$7y - 10 = -4$$, so $$7y = 6$$, and $$y = 6/7$$. This contradicts $$y \ge 3/2$$.
Case 4: $$2y - 3 < 0$$ and $$5y - 7 \ge 0$$, which means $$y < 3/2$$ and $$y \ge 7/5$$. This is impossible.
So, we have $$y = \operatorname{tg} x = 8/3$$ or $$y = \operatorname{tg} x = 0$$.
For $$\operatorname{tg} x = 8/3$$, $$x = \operatorname{arctg}(8/3) + \pi n$$, where $$n \in \mathbb{Z}$$.
For $$\operatorname{tg} x = 0$$, $$x = \pi k$$, where $$k \in \mathbb{Z}$$.
The roots are $$x = \pi k$$ and $$x = \operatorname{arctg}(8/3) + \pi n$$, where $$k, n \in \mathbb{Z}$$.