The number of distinct of real roots of the equation `tan^(2)2x+2tan2x tan3x-1=0` in the interval `[0,(pi)/(2)]` is

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the equation \( \tan^2(2x) + 2\tan(2x)\tan(3x) - 1 = 0 \) in the interval \( [0, \frac{\pi}{2}] \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \tan^2(2x) + 2\tan(2x)\tan(3x) - 1 = 0 \]
We can rearrange this to:
\[ \tan^2(2x) + 2\tan(2x)\tan(3x) = 1 \]

### Step 2: Use the identity for tangent addition
We can use the identity for the tangent of a sum:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Let \( A = 2x \) and \( B = 3x \). Thus, we can express the left-hand side as:
\[ \tan(2x + 3x) = \tan(5x) \]
So, we rewrite the equation as:
\[ \tan(2x)\tan(5x) = 1 \]

### Step 3: Set up the equation
This implies:
\[ \tan(2x) \tan(5x) = 1 \]
This can be rewritten as:
\[ \tan(2x) = \cot(5x) \]
or
\[ \tan(2x) = \frac{1}{\tan(5x)} \]

### Step 4: Solve the equation
From the identity \( \tan A = \cot B \), we have:
\[ \tan(2x) = \tan\left(\frac{\pi}{2} - 5x\right) \]
This leads to two cases:
1. \( 2x = \frac{\pi}{2} - 5x + n\pi \) for some integer \( n \)
2. \( 2x = \frac{\pi}{2} - 5x + n\pi \)

### Step 5: Solve for \( x \)
From the first case:
\[ 2x + 5x = \frac{\pi}{2} + n\pi \]
\[ 7x = \frac{\pi}{2} + n\pi \]
\[ x = \frac{\pi}{14} + \frac{n\pi}{7} \]

### Step 6: Find distinct roots in the interval
Now we need to find the values of \( n \) such that \( x \) is in \( [0, \frac{\pi}{2}] \):
- For \( n = 0 \):
\[ x = \frac{\pi}{14} \approx 0.224 \text{ (valid)} \]
- For \( n = 1 \):
\[ x = \frac{\pi}{14} + \frac{\pi}{7} = \frac{3\pi}{14} \approx 0.673 \text{ (valid)} \]
- For \( n = 2 \):
\[ x = \frac{\pi}{14} + \frac{2\pi}{7} = \frac{5\pi}{14} \approx 1.107 \text{ (not valid)} \]

Thus, the valid values of \( n \) are \( 0 \) and \( 1 \), giving us two distinct roots in the interval \( [0, \frac{\pi}{2}] \).

### Final Answer
The number of distinct real roots of the equation in the interval \( [0, \frac{\pi}{2}] \) is **2**.