Question

If \( \sin^{-1}(4x) - \cos^{-1}(3x) = \frac{\pi}{6} \), then \( x = \):

• A. \( \frac{\sqrt{3}}{2\sqrt{7}} \)
• B. \( \frac{\sqrt{3}}{4\sqrt{7}} \)
• C. \( \frac{\sqrt{3}}{2\sqrt{13}} \)
• D. \( \frac{\sqrt{3}}{4\sqrt{13}} \)

Hint:

When solving inverse trigonometric equations, use known identities for sums of inverse functions and verify the solutions by substituting into the original equation.

Answer 1

The Correct Option is C

We are given the equation: \[ \sin^{-1}(4x) - \cos^{-1}(3x) = \frac{\pi}{6}. \] We need to find the value of \( x \). 

Step 1: First, use the identity \( \cos^{-1} \theta = \frac{\pi}{2} - \sin^{-1} \theta \). Substituting this into the equation, we get: \[ \cos^{-1}(3x) = \frac{\pi}{2} - \sin^{-1}(3x). \] Now substitute this into the original equation: \[ \sin^{-1}(4x) - \left( \frac{\pi}{2} - \sin^{-1}(3x) \right) = \frac{\pi}{6}. \] Simplifying the equation: \[ \sin^{-1}(4x) - \frac{\pi}{2} + \sin^{-1}(3x) = \frac{\pi}{6}. \] \

Step 2: Now, rearrange the equation: \[ \sin^{-1}(4x) + \sin^{-1}(3x) = \frac{\pi}{6} + \frac{\pi}{2}. \] Simplifying the right-hand side: \[ \sin^{-1}(4x) + \sin^{-1}(3x) = \frac{2\pi}{3}. \] 

Step 3: We now use the identity for the sum of two inverse sine functions: \[ \sin^{-1} a + \sin^{-1} b = \sin^{-1} \left( a \sqrt{1 - b^2} + b \sqrt{1 - a^2} \right). \] Let \( a = 4x \) and \( b = 3x \), so the equation becomes: \[ \sin^{-1}(4x) + \sin^{-1}(3x) = \sin^{-1} \left( 4x \sqrt{1 - (3x)^2} + 3x \sqrt{1 - (4x)^2} \right). \] Since we know that the left-hand side equals \( \frac{2\pi}{3} \), we have: \[ \sin^{-1} \left( 4x \sqrt{1 - 9x^2} + 3x \sqrt{1 - 16x^2} \right) = \frac{2\pi}{3}. \] We now solve for \( x \). 

Step 4: Instead of solving the complicated trigonometric terms directly, we try the options. Let’s check option (3), \( x = \frac{\sqrt{3}}{2\sqrt{13}} \). Substitute \( x = \frac{\sqrt{3}}{2\sqrt{13}} \) into the equation \( \sin^{-1}(4x) - \cos^{-1}(3x) = \frac{\pi}{6} \): \[ 4x = 4 \times \frac{\sqrt{3}}{2\sqrt{13}} = \frac{2\sqrt{3}}{\sqrt{13}}, \quad 3x = 3 \times \frac{\sqrt{3}}{2\sqrt{13}} = \frac{3\sqrt{3}}{2\sqrt{13}}. \] Now we substitute these into the sine and cosine inverse functions. First, we compute \( \sin^{-1}(4x) \) and \( \cos^{-1}(3x) \). \[ \sin^{-1}(4x) = \sin^{-1} \left( \frac{2\sqrt{3}}{\sqrt{13}} \right), \quad \cos^{-1}(3x) = \cos^{-1} \left( \frac{3\sqrt{3}}{2\sqrt{13}} \right). \] After calculating both terms, we find that: \[ \sin^{-1}(4x) - \cos^{-1}(3x) = \frac{\pi}{6}. \] Thus, the value of \( x \) is: \[ \boxed{\frac{\sqrt{3}}{2\sqrt{13}}}. \]