if `tan^(-1)(2x)+tan^(-1)(3x)=npi+(pi)/(4),nepsilonI` then number of order pair (s) `(n,x)` is (are)`

Explanation

To solve the equation \( \tan^{-1}(2x) + \tan^{-1}(3x) = n\pi + \frac{\pi}{4} \), we will follow these steps:

### Step 1: Use the addition formula for inverse tangent
We know that:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
provided \( ab < 1 \).

Applying this to our equation:
\[ \tan^{-1}(2x) + \tan^{-1}(3x) = \tan^{-1}\left(\frac{2x + 3x}{1 - 2x \cdot 3x}\right) = \tan^{-1}\left(\frac{5x}{1 - 6x^2}\right) \]

### Step 2: Set the equation
Now we can rewrite the equation as:
\[ \tan^{-1}\left(\frac{5x}{1 - 6x^2}\right) = n\pi + \frac{\pi}{4} \]

### Step 3: Remove the inverse tangent
Taking the tangent of both sides gives us:
\[ \frac{5x}{1 - 6x^2} = \tan\left(n\pi + \frac{\pi}{4}\right) \]
Since \( \tan(n\pi + \frac{\pi}{4}) = \tan\left(\frac{\pi}{4}\right) = 1 \), we have:
\[ \frac{5x}{1 - 6x^2} = 1 \]

### Step 4: Cross-multiply and rearrange
Cross-multiplying gives:
\[ 5x = 1 - 6x^2 \]
Rearranging this equation results in:
\[ 6x^2 + 5x - 1 = 0 \]

### Step 5: Solve the quadratic equation
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 6, b = 5, c = -1 \):
\[ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 6 \cdot (-1)}}{2 \cdot 6} \]
\[ x = \frac{-5 \pm \sqrt{25 + 24}}{12} \]
\[ x = \frac{-5 \pm \sqrt{49}}{12} \]
\[ x = \frac{-5 \pm 7}{12} \]

Calculating the two possible values for \( x \):
1. \( x = \frac{2}{12} = \frac{1}{6} \)
2. \( x = \frac{-12}{12} = -1 \)

### Step 6: Find corresponding \( n \) values
Now we substitute \( x = -1 \) and \( x = \frac{1}{6} \) back into the original equation to find \( n \).

1. For \( x = -1 \):
\[ n\pi + \frac{\pi}{4} = \tan^{-1}\left(\frac{5(-1)}{1 - 6(-1)^2}\right) = \tan^{-1}\left(\frac{-5}{1 - 6}\right) = \tan^{-1}(5) \]
This gives:
\[ n\pi + \frac{\pi}{4} = \frac{\pi}{4} \implies n = 0 \]
So, one ordered pair is \( (0, -1) \).

2. For \( x = \frac{1}{6} \):
\[ n\pi + \frac{\pi}{4} = \tan^{-1}\left(\frac{5 \cdot \frac{1}{6}}{1 - 6 \cdot \left(\frac{1}{6}\right)^2}\right) = \tan^{-1}\left(\frac{\frac{5}{6}}{1 - \frac{1}{6}}\right) = \tan^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right) = \tan^{-1}(1) \]
This gives:
\[ n\pi + \frac{\pi}{4} = \frac{\pi}{4} \implies n = 0 \]
So, the second ordered pair is \( (0, \frac{1}{6}) \).

### Conclusion
The number of ordered pairs \( (n, x) \) is \( 2 \):
1. \( (0, -1) \)
2. \( (0, \frac{1}{6}) \)