Find $\;$ $x$ $\;$ if $\;$ $\sin^{-1} \left(1 - x\right) - 2 \sin^{-1} \left(x\right) = \dfrac{\pi}{2}$
Let $\;$ $\sin^{-1} \left(1 - x\right) = \alpha$, $\;$ $\sin^{-1} \left(x\right) = \beta$ $\;\;\; \cdots \; (1)$
Then, $\;$ $\sin \left(\alpha\right) = 1 - x$, $\;$ $\sin \left(\beta\right) = x$ $\;\;\; \cdots \; (2)$
Given: $\;$ $\sin^{-1} \left(1 - x\right) - 2 \sin^{-1} \left(x\right) = \dfrac{\pi}{2}$
i.e. $\;$ $\alpha - 2 \beta = \dfrac{\pi}{2}$ $\;$ [by equation $(1)$]
i.e. $\;$ $\alpha = \dfrac{\pi}{2} + 2 \beta$
$\therefore \;$ $\sin \left(\alpha\right) = \sin \left(\dfrac{\pi}{2} + 2 \beta\right)$
i.e. $\;$ $\sin \left(\alpha\right) = \cos \left(2 \beta\right)$
i.e. $\;$ $\sin \left(\alpha\right) = 1 - 2 \sin^2 \left(\beta\right)$
i.e. $\;$ $1 - x = 1 - 2 x^2$ $\;$ [by equation $(2)$]
i.e. $\;$ $2 x^2 - x = 0$
i.e. $\;$ $x \left(2 x - 1\right) = 0$
$\implies$ $x = 0$ $\;\;\;$ OR $\;\;\;$ $x = \dfrac{1}{2}$
Putting $\;$ $x = 0$ $\;$ in the given equation,
$\begin{aligned}
LHS & = \sin^{-1} \left(1 - 0\right) - 2 \sin^{-1} \left(0\right) \\\\
& = \sin^{-1} \left(1\right) - 0 \\\\
& = \dfrac{\pi}{2} = RHS
\end{aligned}$
Putting $\;$ $x = \dfrac{1}{2}$ $\;$ in the given equation,
$\begin{aligned}
LHS & = \sin^{-1} \left(1 - \dfrac{1}{2}\right) - 2 \sin^{-1} \left(\dfrac{1}{2}\right) \\\\
& = \sin^{-1} \left(\dfrac{1}{2}\right) - 2 \sin^{-1} \left(\dfrac{1}{2}\right) \\\\
& = - \sin^{-1}\left(\dfrac{1}{2}\right) \\\\
& = - \dfrac{\pi}{6} \neq RHS
\end{aligned}$
$\therefore \;$ The solution set is $\left\{0 \right\}$