Find the roots of the equation ` cot x - cos x=1-cot x cos x `

Answer

To find the roots of the equation \( \cot x - \cos x = 1 - \cot x \cos x \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cot x - \cos x = 1 - \cot x \cos x \]
Rearranging this gives:
\[ \cot x - 1 - \cos x + \cot x \cos x = 0 \]

### Step 2: Factor the equation
We can factor this expression:
\[ \cot x - 1 + \cos x (\cot x - 1) = 0 \]
This can be rewritten as:
\[ (\cot x - 1)(1 + \cos x) = 0 \]

### Step 3: Set each factor to zero
Now we set each factor to zero:
1. \( \cot x - 1 = 0 \)
2. \( 1 + \cos x = 0 \)

### Step 4: Solve the first equation
From \( \cot x - 1 = 0 \), we have:
\[ \cot x = 1 \]
This implies:
\[ x = \frac{\pi}{4} + n\pi \quad \text{(where \( n \) is any integer)} \]

### Step 5: Solve the second equation
From \( 1 + \cos x = 0 \), we have:
\[ \cos x = -1 \]
This gives:
\[ x = \pi + 2n\pi \quad \text{(where \( n \) is any integer)} \]

### Step 6: Combine the solutions
The solutions to the original equation are:
\[ x = \frac{\pi}{4} + n\pi \quad \text{and} \quad x = \pi + 2n\pi \]

### Final Answer
Thus, the roots of the equation \( \cot x - \cos x = 1 - \cot x \cos x \) are:
\[ x = \frac{\pi}{4} + n\pi \quad \text{and} \quad x = \pi + 2n\pi \]