Question

4. For \( x \in(-1,1] \), the number of solutions of the equation \( \sin ^{-1} x=2 \tan ^{-1} x \) is equal to \( \qquad \) [JEE (Main)-2023]

Answer 1

\(sin^{-1}x = 2tan^{-1}x\)

\(sin ^{-1}x = sin ^{-1} (\frac {2x}{1+x^2})\)

\(x = \frac {2x}{1+x^2}\)

x(1 + x²) = 2x

x + x³ = 2x

x³ - x = 0

Hence 0, -1, 1 are

solution of equation

i.e. x(x² - 12) = 0

x(x + 1)(x - 1) = 0

x = 0, x + 1 = 0, x - 1 = 0

 x = -1, x = 1

\(\therefore\) It has 3 solutions.