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.

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.
