<s>[INST] Why did this extraneous root creep into the solution? I was solving this equation and proceeded as follows:
$$\arcsin (1-x) - 2\arcsin (x) = \frac{π}{2}$$
$$\implies \arcsin(1-x) = \frac{π}{2} + 2\arcsin (x)$$
$$\implies \sin (\arcsin (1-x)) = \sin \left( \frac {π}{2} + 2\arcsin (x)\right)$$
 $$\implies (1-x) = \cos \left(2\arcsin(x)\right)$$
 $$\require{cancel}\implies \cancel{1}-x =\cancel{1}- 2 \left(\sin\arcsin (x)\right)^2$$
 $$\implies -x = -2x^2$$
$$\therefore x = 0$$
 Or $$ x=\frac{1}{2}$$
However, $x=\frac{1}{2}$ doesn't satisfy the original equation.
I understand that extraneous roots do creep in while solving inverse trig problems, but I wonder why it crept in here.
If possible (if it doesn't make the question too broad), I'd like to know the general causes for the occurrence of extraneous roots in inverse trig equations, too.
 [/INST] When $x=\frac12$, you have $\arcsin(x)=\frac{\pi}{6}$ and $\arcsin(1-x)=\frac{\pi}{6}$ 
so  $\sin (\arcsin (1-x))=\sin \left( \frac {π}{6} \right)$ and $\sin \left( \frac {π}{2} + 2\arcsin (x)\right) =\sin \left( \frac {5π}{6} \right) $ 
showing your third line would be a correct equality when $x=\frac12$ since  $\sin \left( \frac {π}{6} \right)=\sin \left( \frac {5π}{6} \right) =\frac12$
but your second line would not be a correct equality when $x=\frac12$ since $\frac {π}{6}\not = \frac {5π}{6}$ 
and it is this use of $\sin$ which creates an equality which was not in the original expression. 
$\arcsin (1-x) - 2\arcsin (x) = \frac{π}{2} \Rightarrow x=0 \text{ or } x=\frac12$ is a correct statement, but checking shows $x=0 \Rightarrow \arcsin (1-x) - 2\arcsin (x) = \frac{π}{2}$ while $x=\frac12 \not \Rightarrow \arcsin (1-x) - 2\arcsin (x) = \frac{π}{2}$
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