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.
 A: The extraneous root is introduced in the step
$$\sin^{-1}(1-\sin y)=\frac{\pi}{2}+2y \implies 
 1-\sin y =\sin (\frac{\pi}{2}+2y)$$
where $y$ in the left equation is restricted to $y\le 0$, while it is not restricted in the right equation.
A: Substitutions like that must be done taking a note of the domain and range.
First check the domain of $x$ in the given equation, that is, after solving two inequalities, $x \in [0,1]$
You substituted $x = \sin (\theta)$ (I prefer it like that)
This implies that $\theta \in \left[0, \frac{\pi}{2}\right]$
$$ \sin^{-1}(1-\sin(\theta))-2\sin^{-1}(\sin(\theta))= \frac{\pi}{2} $$
Which simplifies to $ sin^{-1}(1-\sin(\theta))= \frac{\pi}{2}+2\theta$, all good so far, except.
LHS has the range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, so RHS must also lie in that range and after solving the inequality $-\frac{\pi}{2}\le \frac{\pi}{2}+2\theta\le \frac{\pi}{2}$
now the range of $\theta$ comes out to be $\left[-\frac{\pi}{2},0\right]$  Which is  clearly conflicting from the prior domain except $\theta=0$ for which $x=0$
You might have just taken sine on both sides after the substitution, which caused the ambiguity.
A: $$2\sin^{-1}x=\sin^{-1}(1-x)-\dfrac\pi2$$
Using Principal values
$$-\dfrac\pi2-\dfrac\pi2\le2\sin^{-1}x\le\dfrac\pi2-\dfrac\pi2$$
$$\iff-\dfrac\pi2\le\sin^{-1}x\le0\iff x\le0$$
A: 
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.
A: An easy way to hunt down where specifically the erroneous solution is introduced is to simply substitute the erroneous solution into each step of the equation, and find the first step where it becomes valid. In this case, it's between the steps
$$\sin^{-1}(1-\sin y)-2y=\frac{\pi}{2} \tag{1}$$
and
$$1-\sin y =\sin \left(\frac{\pi}{2}+2y\right). \tag{2}$$
We are letting $y = \sin^{-1} x$ (in which case we get $\sin^{-1}(\sin(y)) = y$ as needed), so let $y = \frac{\pi}{6}$. Plugging into $(1)$, we get
$$\sin^{-1}\left(1 - \sin\left(\frac{\pi}{6}\right)\right) - 2\frac{\pi}{6} = \sin^{-1}\left(\frac{1}{2}\right) - \frac{\pi}{3} = -\frac{\pi}{6} \neq \frac{\pi}{2},$$
hence $(1)$ is not satisfied, but plugging into $(2)$,
$$1 - \sin(y) = \frac{1}{2} = \sin\left(\frac{5\pi}{6}\right) = \sin\left(\frac{\pi}{2} + 2\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{2} + 2y\right),$$
i.e. $(2)$ is satisfied. So, we've established that this step is where the erroneous solution is created.
As in my comment, the flow of logic is not reversible here. It is the case that $(1) \implies (2)$, but it's not the case that $(2) \implies (1)$.
To get from $(2)$ to $(1)$, you add $2y$ to both sides, then take $\sin$ of both sides, the latter operation being the one that can't be reversed. The left hand side, after adding $2y$, becomes
$$\sin^{-1}(1 - \sin y).$$
Taking $\sin$ of this yields
$$\sin(\sin^{-1}(1 - \sin y)) = 1 - \sin y.$$
This may seem obvious, but you need to be careful, since $\sin^{-1}$ is not a true inverse of $\sin$. It is true that $\sin(\sin^{-1}(a)) = a$ for any $a \in [-1, 1]$, but it's not true that $\sin^{-1}(\sin(\theta)) = \theta$ for any $\theta \in \Bbb{R}$ (or even, indeed, $\theta \in [0, 2\pi)$). So, if you were to try to reverse this step, to go from $(2)$ back to $(1)$ by taking $\sin^{-1}$ of both sides, you'll get
$$\sin^{-1}(1 - \sin(y)) = \sin^{-1}\left(\sin\left(\frac{\pi}{2} + 2y\right)\right),$$
and the right hand side will not necessarily simplify back down to the expression $\frac{\pi}{2} + 2y$ that we need.
What's an example of this phenomenon? Well, we have one, thanks to the erroneous solution. In this case, taken from the calculations above, we have $\sin\left(\frac{\pi}{6}\right) = \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}$, and so
$$\sin^{-1}\left(\sin\left(\frac{5\pi}{6}\right)\right) = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}.$$
So, we've established that the obvious method of reversing the step simply doesn't take: you cannot take $\sin^{-1}$ of both sides and expect everything to cancel back to how it was. The presence of the erroneous solution tells us that we were doomed from the start; there's no way to go from $(2)$ to $(1)$ logically, simply because $(2)$ has more solutions than $(1)$.
Ultimately, it's not a big deal. Most methods for solving equations (linear/matrix equations being something of an exception) do this same thing, where each step implies the previous, but not necessarily the other way around. It means that, once we get our list of solutions, we must check to see which ones are true solutions and which are erroneous. Most of the time, we get back a finite list of possible solutions, so checking is not too onerous. Most of the rest of the time, we might get back an infinite family of possible solutions indexed by some parameter(s), and plugging in these possible solutions will quickly reveal which possible solutions are erroneous, and which are not. Even if you think your method will have no erroneous solutions, it never hurts to plug them in to check, just as way of verification that no error occurred.
Anyway, I hope that helps.
