Solve ` 2 cot 2x -3 cot 3x = tan 2x ` .

Answer

To solve the equation \( 2 \cot 2x - 3 \cot 3x = \tan 2x \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ 2 \cot 2x - 3 \cot 3x = \tan 2x \]

### Step 2: Move \( \tan 2x \) to the left side
We can rewrite the equation as:
\[ 2 \cot 2x - 3 \cot 3x - \tan 2x = 0 \]

### Step 3: Express \( \tan 2x \) in terms of sine and cosine
Recall that:
\[ \tan 2x = \frac{\sin 2x}{\cos 2x} \]
So we can rewrite the equation as:
\[ 2 \cot 2x - 3 \cot 3x - \frac{\sin 2x}{\cos 2x} = 0 \]

### Step 4: Express cotangent in terms of sine and cosine
Using the identity \( \cot x = \frac{\cos x}{\sin x} \), we can express \( \cot 2x \) and \( \cot 3x \):
\[ 2 \frac{\cos 2x}{\sin 2x} - 3 \frac{\cos 3x}{\sin 3x} - \frac{\sin 2x}{\cos 2x} = 0 \]

### Step 5: Multiply through by \( \sin 2x \sin 3x \cos 2x \) to eliminate denominators
This gives us:
\[ 2 \cos 2x \sin 3x \cos 2x - 3 \cos 3x \sin 2x \cos 2x - \sin^2 2x \sin 3x = 0 \]

### Step 6: Rearranging terms
Now we can rearrange the equation:
\[ 2 \cos 2x \sin 3x - 3 \cos 3x \sin 2x - \sin^2 2x = 0 \]

### Step 7: Use the sine subtraction formula
Using the identity \( \sin A - \sin B = 2 \cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right) \):
\[ \sin 3x - 2x = 0 \]

### Step 8: Solve for \( x \)
We can set:
\[ 3x - 2x = n\pi \quad \text{(where \( n \) is an integer)} \]
This simplifies to:
\[ x = n\pi \]

### Step 9: Consider \( \sin x = 0 \)
From the equation \( \sin x = 0 \), we have:
\[ x = n\pi \quad \text{(where \( n \) is any integer)} \]

### Final Solution
The solutions to the equation \( 2 \cot 2x - 3 \cot 3x = \tan 2x \) are:
\[ x = n\pi \quad \text{for } n \in \mathbb{Z} \]