tg x cos x + tg x = cos x+1

To solve the equation \( \tan x \cos x + \tan x = \cos x + 1 \), we can start by rewriting the equation using the identity for \(\tan x\):

\[ \tan x = \frac{\sin x}{\cos x} \]

Thus, we can substitute \(\tan x\) into the equation:

\[ \frac{\sin x}{\cos x} \cos x + \frac{\sin x}{\cos x} = \cos x + 1 \]

This simplifies to:

\[ \sin x + \frac{\sin x}{\cos x} = \cos x + 1 \]

We multiply through by \(\cos x\) (keeping in mind that \(\cos x \neq 0\)):

\[ \sin x \cos x + \sin x = \cos^2 x + \cos x \]

Now we can combine terms on the left-hand side:

\[ \sin x (\cos x + 1) = \cos^2 x + \cos x \]

Next, we can rearrange this:

\[ \sin x (\cos x + 1) - (\cos^2 x + \cos x) = 0 \]

Factoring out \(\cos x\) from the terms on the right, we obtain:

\[ \sin x (\cos x + 1) - \cos x(\cos x + 1) = 0 \]

Factoring the entire expression gives us:

\[ (\cos x + 1)(\sin x - \cos x) = 0 \]

This gives us two cases to solve:

1. \( \cos x + 1 = 0 \)
2. \( \sin x - \cos x = 0 \)

For the first case:

\[ \cos x + 1 = 0 \implies \cos x = -1 \]

This occurs at:

\[ x = \pi + 2k\pi \quad (k \in \mathbb{Z}) \]

For the second case:

\[ \sin x - \cos x = 0 \implies \sin x = \cos x \]

This happens when:

\[ \tan x = 1 \implies x = \frac{\pi}{4} + k\pi \quad (k \in \mathbb{Z}) \]

Thus, the complete solution set for the equation \( \tan x \cos x + \tan x = \cos x + 1 \) is:

\[ x = \pi + 2k\pi \quad (k \in \mathbb{Z}) \quad \text{and} \quad x = \frac{\pi}{4} + k\pi \quad (k \in \mathbb{Z}) \]