Question

If \( \sin^{-1} x + \cos^{-1} 2x = \frac{\pi}{6} \), then the value of \( x \) is

• A. \( \frac{1}{2} \)
• B. \( \frac{\sqrt{3}}{2} \)
• C. \( \sqrt{3} \)
• D. \( 1 \)
• E. \( \sqrt{2} \)

Hint:

Use trig identities and convert inverse functions into angles for simplification.

Answer 1

The Correct Option is A

Concept: \[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]

Step 1: Let \( \sin^{-1} x = \theta \). Then: \[ \cos^{-1} 2x = \frac{\pi}{6} - \theta \]

Step 2: Take cosine. \[ 2x = \cos\left(\frac{\pi}{6} - \theta\right) \] Using identity: \[ \cos(A-B) = \cos A \cos B + \sin A \sin B \] \[ 2x = \frac{\sqrt{3}}{2}\cos\theta + \frac{1}{2}\sin\theta \] \[ = \frac{\sqrt{3}}{2}\sqrt{1-x^2} + \frac{1}{2}x \]

Step 3: Solve equation. Multiply by 2: \[ 4x = \sqrt{3}\sqrt{1-x^2} + x \] \[ 3x = \sqrt{3}\sqrt{1-x^2} \] Square: \[ 9x^2 = 3(1-x^2) \] \[ 9x^2 = 3 - 3x^2 \Rightarrow 12x^2 = 3 \Rightarrow x^2 = \frac{1}{4} \] \[ x = \frac{1}{2} (\text{valid in domain}) \]