Problem
Solve: \( \tan^{-1}(2x) + \tan^{-1}(3x) = n\pi + \frac{3\pi}{4} \)
Solution
Let:
\[ A = \tan^{-1}(2x), \quad B = \tan^{-1}(3x) \]
Step 1: Use tan(A + B)
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \]
\[ = \frac{2x + 3x}{1 – 6x^2} = \frac{5x}{1 – 6x^2} \]
Step 2: Use given equation
\[ \tan(A + B) = \tan\left(n\pi + \frac{3\pi}{4}\right) \]
\[ = \tan\left(\frac{3\pi}{4}\right) = -1 \]
Step 3: Form equation
\[ \frac{5x}{1 – 6x^2} = -1 \]
Step 4: Solve
\[ 5x = -(1 – 6x^2) \]
\[ 6x^2 – 5x – 1 = 0 \]
\[ (3x – 1)(2x + 1) = 0 \]
Step 5: Final values
\[ x = \frac{1}{3}, \quad x = -\frac{1}{2} \]
Final Answer
\[ \boxed{x = \frac{1}{3},\; -\frac{1}{2}} \]
Explanation
Using tan(A+B) identity and solving the resulting quadratic equation.