Question

For x ∈ (–1, 1], the number of solutions of the equation sin^-1x = 2 tan^-1 x is equal to____

Answer 1

Correct Answer: 2

We are given the equation \( \sin^{-1} x = 2 \tan^{-1} x \). First, apply the inverse functions: \[ \sin^{-1} x = \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \] So, \( x = \frac{2x}{1 + x^2} \), simplifying to: \[ x(1 + x^2) = 2x \] \[ x + x^3 = 2x \] \[ x^3 = x \] \[ x(x^2 - 1) = 0 \] Thus, \( x = 0, 1, -1 \). However, the value \( x = 1 \) and \( x = -1 \) are not within the domain, leaving us with only 2 solutions: \( x = 0 \) and \( x = -1 \). Thus, the number of solutions is 2.