1) 3tg x + 2tg x - 1 = 0; 2) 2ctg x + 3ctg x - 2 = 0;

To solve the equations involving the tangent (\( \tan \)) and cotangent (\( \cot \)) functions, let's analyze them one by one.

1. \( 3\tan x + 2\tan x - 1 = 0 \)

First, combine like terms:

\[ 5\tan x - 1 = 0 \]

Now, isolate \( \tan x \):

\[ 5\tan x = 1 \]

\[ \tan x = \frac{1}{5} \]

Now, we find the general solution for \( x \):

\[ x = \arctan\left(\frac{1}{5}\right) + n\pi \quad \text{(where n is any integer)}\]

2. \( 2\cot x + 3\cot x - 2 = 0 \)

Similarly, combine like terms:

\[ 5\cot x - 2 = 0 \]

Isolate \( \cot x \):

\[ 5\cot x = 2 \]

\[ \cot x = \frac{2}{5} \]

Finding the general solution for \( x \):

Recall that \( \cot x = \frac{1}{\tan x} \), thus:

\[ \tan x = \frac{5}{2} \]

Now the general solution is:

\[ x = \arctan\left(\frac{5}{2}\right) + n\pi \quad \text{(where n is any integer)}\]

Summary of Solutions:

1. For \( 3\tan x + 2\tan x - 1 = 0 \):

   • \( x = \arctan\left(\frac{1}{5}\right) + n\pi, \, n \in \mathbb{Z} \)

2. For \( 2\cot x + 3\cot x - 2 = 0 \):

   • \( x = \arctan\left(\frac{5}{2}\right) + n\pi, \, n \in \mathbb{Z} \)