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