Why is this solution excluded from the solution set?

Question

There is a question says :

Solve the equations : $\operatorname{arcsin}2x + \operatorname{arccos} x = \frac{\pi}6$

I ended up with solutions $+ \frac12$ or $-\frac12$ .. and the correct answer was only $-\frac12$

I know when we substitute in the original equation the right answer will be only $-\frac12$

But is there another reason?

Answer 1

As the principal value of $\displaystyle \arccos x$ lies in $\displaystyle [0,\pi]$ and $\displaystyle \arccos\frac12=\frac\pi3$

For $\displaystyle x=\frac12,\arcsin2x=\arcsin1=\frac\pi2$ as the principal value of $\displaystyle \arcsin y$ lies in $\left[-\frac\pi2,\frac\pi2\right]$

So, $\displaystyle\arcsin\left(2\cdot\frac12\right)+\arccos\frac12=\cdots=\frac{5\pi}6$

The extraneous root comes as $\displaystyle\sin(\pi-A)=\sin A$

Here $\displaystyle A=\frac{5\pi}6$