Aspire Faculty ID #18108 · Topic: JEE Main 2026 (28 January Morning Shift) · Just now
JEE Main 2026 (28 January Morning Shift)

If $ k = \tan\left( \frac{\pi}{4} + \frac{1}{2}\cos^{-1}\left(\frac{2}{3}\right) \right) + \tan\left( \frac{1}{2}\sin^{-1}\left(\frac{2}{3}\right) \right) $, then the number of solutions of the equation

$ \sin^{-1}(kx - 1) = \sin^{-1}x - \cos^{-1}x $ is ______

Solution

Let $ \theta = \frac{1}{2}\sin^{-1}\left(\frac{2}{3}\right) $

$ \Rightarrow k = \tan\theta + \cot\theta = \frac{1}{\sin\theta\cos\theta} = \frac{2}{\sin 2\theta} $

$ k = 3 $

$ \sin^{-1}(3x - 1) = \sin^{-1}x - \cos^{-1}x $

$ \sin^{-1}(3x - 1) = \frac{\pi}{2} - 2\cos^{-1}x $

$ 3x - 1 = \sin\left( \frac{\pi}{2} - 2\cos^{-1}x \right) $

$ \Rightarrow 3x - 1 = 2x^2 - 1 $

$ \Rightarrow x = 0,; \frac{3}{2} \text{ (rejected)} $

No. of solution $ = 1 $
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