`tan^(-1) 2x+tan^(-1) 3x=npi+(3pi)/4 , " then " x=`

Options

A. `1 , -1/6`
B. `1/2 , 1/3`
C. `4,5`
D. none

Answer

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

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

### Step 1: Apply the formula
Let \( a = 2x \) and \( b = 3x \). Then, we can write:

\[ \tan^{-1}(2x) + \tan^{-1}(3x) = \tan^{-1}\left(\frac{2x + 3x}{1 - (2x)(3x)}\right) = \tan^{-1}\left(\frac{5x}{1 - 6x^2}\right) \]

### Step 2: Set the equation
Now, we set this equal to \( n\pi + \frac{3\pi}{4} \):

\[ \tan^{-1}\left(\frac{5x}{1 - 6x^2}\right) = n\pi + \frac{3\pi}{4} \]

### Step 3: Take the tangent of both sides
Taking the tangent of both sides gives us:

\[ \frac{5x}{1 - 6x^2} = \tan\left(n\pi + \frac{3\pi}{4}\right) \]

Since \( \tan(n\pi + \theta) = \tan(\theta) \), we have:

\[ \tan\left(n\pi + \frac{3\pi}{4}\right) = \tan\left(\frac{3\pi}{4}\right) = -1 \]

Thus, we can write:

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

### Step 4: Cross-multiply and rearrange
Cross-multiplying gives us:

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

This simplifies to:

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

Rearranging the equation:

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

### Step 5: Solve the quadratic equation
Now, we will use the quadratic formula to solve for \( x \):

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

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

\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 6 \cdot (-1)}}{2 \cdot 6} \]

Calculating the discriminant:

\[ x = \frac{5 \pm \sqrt{25 + 24}}{12} = \frac{5 \pm \sqrt{49}}{12} = \frac{5 \pm 7}{12} \]

### Step 6: Find the values of \( x \)
This gives us two possible solutions:

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

### Final Answer
Thus, the values of \( x \) are:

\[ x = 1 \quad \text{or} \quad x = -\frac{1}{6} \]