If `tan^(-1)(2x)+tan^(-1)(3x)=(pi)/(4)`, then find the value of x.

Answer

To solve the equation \( \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4} \), we can use the formula for the sum of inverse tangents:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]

where \( a = 2x \) and \( b = 3x \).

### Step 1: Apply the formula
Using the formula, we have:

\[ \tan^{-1}(2x) + \tan^{-1}(3x) = \tan^{-1}\left(\frac{2x + 3x}{1 - (2x)(3x)}\right) \]

This simplifies to:

\[ \tan^{-1}\left(\frac{5x}{1 - 6x^2}\right) \]

### Step 2: Set the equation equal to \( \frac{\pi}{4} \)
Since we know that \( \tan\left(\frac{\pi}{4}\right) = 1 \), we can set the argument of the tangent inverse equal to 1:

\[ \frac{5x}{1 - 6x^2} = 1 \]

### Step 3: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:

\[ 5x = 1 - 6x^2 \]

### Step 4: Rearrange the equation
Rearranging the equation results in:

\[ 6x^2 + 5x - 1 = 0 \]

### Step 5: Solve the quadratic equation
Now we can solve the quadratic equation using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 6 \), \( b = 5 \), and \( c = -1 \).

Calculating the discriminant:

\[ b^2 - 4ac = 5^2 - 4 \cdot 6 \cdot (-1) = 25 + 24 = 49 \]

Now substituting into the quadratic formula:

\[ x = \frac{-5 \pm \sqrt{49}}{2 \cdot 6} = \frac{-5 \pm 7}{12} \]

Calculating the two possible values for \( x \):

1. \( x = \frac{2}{12} = \frac{1}{6} \)
2. \( x = \frac{-12}{12} = -1 \)

### Step 6: Choose the valid solution
Since \( x \) must be a non-negative value in this context, we discard \( x = -1 \).

Thus, the value of \( x \) is:

\[ \boxed{\frac{1}{6}} \]