If 2sin^(-1)x=-sin^(-1)(2x*sqrt(1-x^(2)))-pi_(i) then x satisfies

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

x \in

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

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

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

\frac{- 1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}

x

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

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