For x ( - 1 , 1 ] , the number of solutions of the equation sin - 1 x = 2 tan - 1 x is equal to ________ . [2023]

Question

For $x \in (- 1, 1],$ the number of solutions of the equation $\sin^{- 1} x = 2 \tan^{- 1} x$ is equal to ________ . [2023]

Smart Achievers · Accepted Answer

(2)

$\sin^{- 1} x = 2 \tan^{- 1} x$

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

$\Rightarrow x (1 + x^{2}) = 2 x \Rightarrow x (1 + x^{2}) - 2 x = 0 \Rightarrow x (x^{2} - 1) = 0$

$\Rightarrow x = 0 or x^{2} - 1 = 0 \Rightarrow x^{2} = 1 \Rightarrow x = \pm 1$ $\Rightarrow x = 0 or x = 1, - 1$

Number of solutions = 2

$∵$ $x = - 1 \notin (- 1, 1]$

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