Find the range of $x$ for which the formula $3 {sin}^{- 1} x = {sin}^{- 1} (3 x - 4 x^{3})$ holds true ?

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Solution

$3 {sin}^{- 1} x = {sin}^{- 1} (3 x - 4 x^{3})$
Putting ${sin}^{- 1} x = A \Rightarrow x = sin A$
$\Rightarrow 3 A = {sin}^{- 1} (3 sin A - 4 {sin}^{3} A)$
$\Rightarrow 3 A = {sin}^{- 1} (sin 3 A) (∵ sin 3 A = 3 sin A - 4 {sin}^{3} A)$
$\Rightarrow 3 A \in [\frac{- π}{2}, \frac{π}{2}] \Rightarrow A \in [\frac{- π}{6}, \frac{π}{6}] \Rightarrow {sin}^{- 1} x \in [\frac{- π}{6}, \frac{π}{6}] \Rightarrow x \in [\frac{- 1}{2}, \frac{1}{2}]$

Therefore, the range of $x$ is $[\frac{- 1}{2}, \frac{1}{2}]$ .

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