How many $\displaystyle \displaystyle x\in\mathbb{R}$ exist such that $\displaystyle \displaystyle \arcsin{x} = \arccos{x} $?
it's just a matter of being familiar with both functions.
$\displaystyle y = \arcsin(x)$ has range $\displaystyle -\frac{\pi}{2} \le y \le \frac{\pi}{2}$
$\displaystyle y = \arccos(x)$ has range $\displaystyle 0 \le y \le \pi$
the ranges intersect in quad I ... the angle in quad I that has the same sine and cosine value is $\displaystyle y = \frac{\pi}{4}$
if you insist on doing it algebraically ...
$\displaystyle \arcsin(x) = \arccos(x)$
$\displaystyle \sin[\arcsin(x)] = \sin[\arccos(x)]$
$\displaystyle x = \sqrt{1 - x^2}$
$\displaystyle x^2 = 1 - x^2$
$\displaystyle 2x^2 - 1 = 0$
$\displaystyle (\sqrt{2} x - 1)(\sqrt{2} x + 1) = 0$
$\displaystyle x = \frac{1}{\sqrt{2}}$ ... the negative solution is invalid since
$\displaystyle y = \arcsin\left(-\frac{1}{\sqrt{2}}\right) = -\frac{\pi}{4}$
and
$\displaystyle y = \arccos\left(-\frac{1}{\sqrt{2}}\right) = \frac{3\pi}{4}$
$\displaystyle \arcsin\left(\frac{1}{\sqrt{2}}\right) = \arccos\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$