sin^-1(1-x)-2sin^-1x=pi/2

$\sin^{- 1} (1 - x) - 2 \sin^{- 1} x = \frac{π}{2} ∴ \sin^{- 1} (1 - x) = \frac{π}{2} + 2 \sin^{- 1} x \Rightarrow (1 - x) = \sin (\frac{π}{2} + 2 \sin^{- 1} x) \Rightarrow (1 - x) = \cos (2 \sin^{- 1} x) \Rightarrow (1 - x) = \cos [\cos^{- 1} (1 - 2 x^{2})] \Rightarrow (1 - x) = 1 - 2 x^{2} \Rightarrow 2 x^{2} - x = 0 \Rightarrow x (2 x - 1) = 0 \Rightarrow x = 0 or 2 x - 1 = 0 \Rightarrow x = 0 or x = \frac{1}{2} For x = \frac{1}{2}, L H S = \sin^{- 1} (1 - x) - 2 \sin^{- 1} x = \sin^{- 1} \frac{1}{2} - 2 \sin^{- 1} \frac{1}{2} = - \sin^{- 1} \frac{1}{2} = - \frac{π}{6}$

≠ RHS

So, $x = \frac{1}{2}$ is not a solution of given equation.

x = 0 satisfy the given equation.

∴ x = 0 is the solution of the given solution.

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sin-1(1-x)-2sin-1x=pi/2

sin^-1(1-x)-2sin^-1x=pi/2