|2 tg x - 3| - |5 tg x - 7| = -4

Let $$y = \tan x$$. The equation becomes $$|2y - 3| - |5y - 7| = -4$$. We consider three cases based on the roots of the absolute value expressions, which are $$y = 3/2$$ and $$y = 7/5$$.
Case 1: $$y < 7/5$$. Then $$2y - 3 < 0$$ and $$5y - 7 < 0$$. The equation is $$-(2y - 3) - (-(5y - 7)) = -4$$, which simplifies to $$-2y + 3 + 5y - 7 = -4$$, so $$3y - 4 = -4$$, giving $$3y = 0$$, so $$y = 0$$. Since $$0 < 7/5$$, this is a valid solution. Thus, $$\tan x = 0$$, which means $$x = n\pi$$ for integer $$n$$.
Case 2: $$7/5 \le y < 3/2$$. Then $$2y - 3 < 0$$ and $$5y - 7 \ge 0$$. The equation is $$-(2y - 3) - (5y - 7) = -4$$, which simplifies to $$-2y + 3 - 5y + 7 = -4$$, so $$-7y + 10 = -4$$, giving $$-7y = -14$$, so $$y = 2$$. However, $$2 \ge 3/2$$, so this case yields no solution.
Case 3: $$y \ge 3/2$$. Then $$2y - 3 \ge 0$$ and $$5y - 7 > 0$$. The equation is $$(2y - 3) - (5y - 7) = -4$$, which simplifies to $$2y - 3 - 5y + 7 = -4$$, so $$-3y + 4 = -4$$, giving $$-3y = -8$$, so $$y = 8/3$$. Since $$8/3 \ge 3/2$$, this is a valid solution. Thus, $$\tan x = 8/3$$, which means $$x = \arctan(8/3) + n\pi$$ for integer $$n$$.
The solutions are $$x = n\pi$$ and $$x = \arctan(8/3) + n\pi$$, where $$n$$ is an integer.