Solve for $x$ in $\sin^{-1}(1-x)-2\sin^{-1}x =\frac{\pi}{2}$ Let $x=\sin y$
$$\sin^{-1}(1-\sin y)-2\sin^{-1}\sin y=\frac{\pi}{2}$$
$$\sin^{-1}(1-\sin y)-2y=\frac{\pi}{2}$$
$$1-\sin y =\sin (\frac{\pi}{2}+2y)$$
$$1-\cos2y=\sin y$$
$$\sin y(2\sin y-1)=0$$
$$x=0,~ \frac 12$$
Clearly, $x=\frac 12$ isn’t correct, because it doesn’t satisfy the original expression
Why was an extraneous root obtained in this solution? I want to know the reason behind it.


Find $x$ satisfying $\sin^{-1}(1-x)-2\sin^{-1}x =\frac{\pi}{2}$.

Common mistakes
While going step-by-step, if you go from

*

*$(\sin^{-1}a)=y\stackrel{\text{to}}{\longrightarrow}\sin(\sin^{-1}a)=a=\sin (y\pm 2n\pi)$, and from

*$\sin y=a\stackrel{\text{to}}{\longrightarrow}\sin^{-1}(\sin y)=\begin{cases}2n\pi+y&y\in\text{I, IV quadrant}\\(2n-1)\pi-y&y\in\text{II, III quadrant}\end{cases}=\sin^{-1}a$,

there occurs an uncertainty$\color{red}{^1}$ of $m\pi$ in $y$. So, the new value of $y$ also begins to satisfy the original equation $(\sin^{-1}a)=y$ or $\sin y=a$.

Your mistakes apart from not doing the domain test
Let $x=\sin y$, then
\begin{align*}
\sin^{-1}(1-\sin y)-2\sin^{-1}\sin y&=\frac{\pi}{2}\\
\end{align*}
$\color{red}{\text{CASE I: $y\in$ I, IV quadrant}}$
\begin{align*}
\sin^{-1}(1-\sin y)-2\color{red}{(y+2n\pi)}&=\frac{\pi}{2}\tag{1}\\
\color{red}{\sin^{-1}(1-\sin y)}&\color{red}{=}\color{red}{\frac{\pi}{2}+2(y+2n\pi)\Rightarrow (y+2n\pi)\in\left[-\frac{\pi}{2},0\right]}\Rightarrow y\in\text{IV quadrant}\tag{2}\\
1-\sin y &=\sin \left(\frac{\pi}{2}+2y\color{red}{+4n\pi}\right)\tag{3}\\
1-\cos2y&=\sin y\\
\sin y(2\sin y-1)&=0\Rightarrow \sin y=0,\ \sin y=\frac 12 
\end{align*}
but $\sin y=\frac 12$ is not allowed by the range test that bounds $y$ in IV quadrant in which sine is negative.
$\color{red}{\text{CASE II: $y\in$ II, III quadrant}}$
\begin{align*}
\sin^{-1}(1-\sin y)-2\color{red}{[(2n-1)\pi-y]}&=\frac{\pi}{2}\tag{4}\\
\color{red}{\sin^{-1}(1-\sin y)}&\color{red}{=}\color{red}{\frac{\pi}{2}+2[(2n-1)\pi-y]\Rightarrow ((2n-1)\pi-y)\in\left[-\frac{\pi}{2},0\right]\Rightarrow y\in\left[(2n-1)\pi,\frac{4n-1}{2}\pi\right]\Rightarrow y\in\text{III quadrant}}\tag{5}\\
1-\sin y &=\sin \left(\frac{\pi}{2}\color{red}{+2[(2n-1)\pi-y]}\right)\tag{6}\\
1-\cos2y&=\sin y\\
\sin y(2\sin y-1)&=0\Rightarrow \sin y=0,\ \sin y=\frac 12 
\end{align*}
but $\sin y=\frac 12$ is not allowed by the range test that bounds $y$ in III quadrant in which sine is negative.
Your mistake apart from the domain test (that gives the answer instantly) is the range test. :p

My solution
Since taking sine both sides produces no uncertainty for both below LHS and RHS can't deviate by even $2\pi$, we get
\begin{align*}
\sin^{-1}(1-x)-\sin^{-1}x&=\frac{\pi}{2}+\sin^{-1}x\\
\sin[\sin^{-1}(1-x)-\sin^{-1}x] &=\sin\left(\frac{\pi}{2}+\sin^{-1}x\right)\\
\sin[\sin^{-1}(1-x)]\cdot\cos[\sin^{-1}x]-\sin(\sin^{-1}x)\cdot\cos[\sin^{-1}(1-x)] &=\cos(\sin^{-1}x)\\
(1-x)\sqrt{1-x^2}-x\sqrt{2x-x^2} &=\sqrt{1-x^2}&\left(\because \sin^{-1}a\in\left[-\frac{\pi}{2},+\frac{\pi}{2}\right]\Rightarrow \cos(\sin^{-1}a)=\color{red}{+}\sqrt{1-a^2}\right)\\
-x\left(\sqrt{1-x^2}+\sqrt{2x-x^2}\right)&=0 &\\
-x\left(\sqrt{1-x^2}+\sqrt{2x-x^2}\right)&=0 &\\
\Rightarrow x&=0&(\because x\in[0,1])
\end{align*}
since both $\sqrt{1-x^2}$ and $\sqrt{2x-x^2}$ can't vanish simultaneously. The former requires $x=\pm 1$ that the latter does not.

$\color{red}{^1}$This occurs due to defining the range of an inverse function as some part of the domain of the original function.
