Solve : `2+tan x. cot\ (x)/(2)+cot x. tan\ (x)/(2)=0`.

Explanation

To solve the equation \( 2 + \tan x \cdot \cot \frac{x}{2} + \cot x \cdot \tan \frac{x}{2} = 0 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ 2 + \tan x \cdot \cot \frac{x}{2} + \cot x \cdot \tan \frac{x}{2} = 0 \]

### Step 2: Substitute variables
Let \( t = \tan x \cdot \cot \frac{x}{2} \). Then, we can rewrite the equation as:
\[ 2 + t + \frac{1}{t} = 0 \]

### Step 3: Rearranging the equation
We can rearrange the equation to isolate \( t \):
\[ t + \frac{1}{t} = -2 \]

### Step 4: Multiply through by \( t \)
Multiply both sides by \( t \) (assuming \( t \neq 0 \)):
\[ t^2 + 1 = -2t \]
Rearranging gives us:
\[ t^2 + 2t + 1 = 0 \]

### Step 5: Factor the quadratic equation
The equation can be factored as:
\[ (t + 1)^2 = 0 \]
Thus, we find:
\[ t + 1 = 0 \quad \Rightarrow \quad t = -1 \]

### Step 6: Substitute back for \( t \)
Recall that \( t = \tan x \cdot \cot \frac{x}{2} \). Therefore, we have:
\[ \tan x \cdot \cot \frac{x}{2} = -1 \]

### Step 7: Rewrite in terms of sine and cosine
Using the definitions of tangent and cotangent:
\[ \tan x = \frac{\sin x}{\cos x} \quad \text{and} \quad \cot \frac{x}{2} = \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} \]
Substituting these into our equation gives:
\[ \frac{\sin x}{\cos x} \cdot \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} = -1 \]

### Step 8: Simplify the equation
This simplifies to:
\[ \frac{\sin x \cdot \cos \frac{x}{2}}{\cos x \cdot \sin \frac{x}{2}} = -1 \]
Cross-multiplying gives:
\[ \sin x \cdot \cos \frac{x}{2} = -\cos x \cdot \sin \frac{x}{2} \]

### Step 9: Use the double angle identity
Using the identity \( \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2} \), we can rewrite the left side:
\[ 2 \sin \frac{x}{2} \cos \frac{x}{2} \cdot \cos \frac{x}{2} = -\cos x \cdot \sin \frac{x}{2} \]
This simplifies to:
\[ 2 \sin \frac{x}{2} \cos^2 \frac{x}{2} = -\cos x \cdot \sin \frac{x}{2} \]

### Step 10: Factor out \( \sin \frac{x}{2} \)
Assuming \( \sin \frac{x}{2} \neq 0 \):
\[ 2 \cos^2 \frac{x}{2} + \cos x = 0 \]

### Step 11: Substitute \( \cos x \)
Using the identity \( \cos x = 2 \cos^2 \frac{x}{2} - 1 \):
\[ 2 \cos^2 \frac{x}{2} + (2 \cos^2 \frac{x}{2} - 1) = 0 \]
This simplifies to:
\[ 4 \cos^2 \frac{x}{2} - 1 = 0 \]

### Step 12: Solve for \( \cos \frac{x}{2} \)
Setting this equal to zero gives:
\[ 4 \cos^2 \frac{x}{2} = 1 \quad \Rightarrow \quad \cos^2 \frac{x}{2} = \frac{1}{4} \]
Taking the square root:
\[ \cos \frac{x}{2} = \pm \frac{1}{2} \]

### Step 13: Find \( \frac{x}{2} \)
The solutions for \( \frac{x}{2} \) are:
\[ \frac{x}{2} = \frac{\pi}{3} + 2n\pi \quad \text{and} \quad \frac{x}{2} = \frac{5\pi}{3} + 2n\pi \]
Thus, multiplying by 2 gives:
\[ x = \frac{2\pi}{3} + 4n\pi \quad \text{and} \quad x = \frac{10\pi}{3} + 4n\pi \]

### Final Solution
The general solutions for \( x \) are:
\[ x = \frac{2\pi}{3} + 4n\pi \quad \text{and} \quad x = \frac{10\pi}{3} + 4n\pi \quad \text{where } n \in \mathbb{Z} \]