Total number of solution of `tan3x-tan2x-tan3xtan2x=1` in `[0,2pi]` is equal to

Options

A. 0
B. 2
C. 4
D. 3

Answer

To solve the equation \( \tan(3x) - \tan(2x) - \tan(3x) \tan(2x) = 1 \) in the interval \([0, 2\pi]\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \tan(3x) - \tan(2x) - \tan(3x) \tan(2x = 1 \]
We can rearrange this to:
\[ \tan(3x) - \tan(2x) = 1 + \tan(3x) \tan(2x) \]

### Step 2: Use the tangent subtraction formula
Using the tangent subtraction formula:
\[ \tan(A) - \tan(B) = \frac{\tan(A) - \tan(B)}{1 + \tan(A)\tan(B)} \]
we can rewrite the left-hand side:
\[ \frac{\tan(3x) - \tan(2x)}{1 + \tan(3x) \tan(2x)} = 1 \]
This simplifies to:
\[ \tan(3x) - \tan(2x) = 1 + \tan(3x) \tan(2x) \]

### Step 3: Set up the new equation
Now we can express this as:
\[ \tan(3x) - \tan(2x) = 1 + \tan(3x) \tan(2x) \]
This implies:
\[ \tan(3x) - \tan(2x) = 1 + \tan(3x) \tan(2x) \]

### Step 4: Solve for \( \tan(3x) \) and \( \tan(2x) \)
Using the identity:
\[ \tan(3x) - \tan(2x) = \tan(3x - 2x) \]
we have:
\[ \tan(3x - 2x) = 1 \]
Thus:
\[ \tan(x) = 1 \]

### Step 5: Find the general solutions
The general solutions for \( \tan(x) = 1 \) are:
\[ x = \frac{\pi}{4} + n\pi \quad \text{for } n \in \mathbb{Z} \]

### Step 6: Determine the specific solutions in the interval \([0, 2\pi]\)
Now we find the specific solutions in the interval \([0, 2\pi]\):
1. For \( n = 0 \):
\[ x = \frac{\pi}{4} \]
2. For \( n = 1 \):
\[ x = \frac{\pi}{4} + \pi = \frac{5\pi}{4} \]

### Step 7: Count the solutions
The solutions \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \) are both within the interval \([0, 2\pi]\).

### Conclusion
Thus, the total number of solutions of the equation \( \tan(3x) - \tan(2x) - \tan(3x) \tan(2x) = 1 \) in the interval \([0, 2\pi]\) is:
\[ \boxed{2} \]