tag removed - is there a solution for equation $arcsin((1-x)^{1/2})=arccos(x^{1/2})$ in which $x$ is rational number

This may look elementary -- but it is most definitely not. This is because there are some nasty branch cuts involved, and making sure that the identity actually holds at all is not easy. The first thing to do is to look at what happens if one expands these functions into their simpler representation using logarithms:
$$ arcsin left( sqrt {1-x} right) =-iln left( sqrt {x}+isqrt {1-x} right) $$
and
$$ arccos left( sqrt {x} right) =1/2pi +iln left( sqrt {1-x}+isqrt {x} right) $$

For essentially all real $x$ outside $(0,1)$, these two quantities are complex. But, as it turns out, there are solutions. The simplest next step is to figure out 'where', and this is best done by splitting into cases. It can be written as
$$cases{i left( -ln left( sqrt {-x}+sqrt {1-x} right) -ln left( sqrt {1-x}-sqrt {-x} right) right) & xleq 0 cr
-1/2pi +arctan left( {frac {sqrt {1-x}}{sqrt {x}}} right) +arctan left( {frac {sqrt {x}}{sqrt {1-x}}} right) & xin(0,1) cr
i left( -ln left( sqrt {x}-sqrt {-1+x} right) -ln left( sqrt {-1+x}+sqrt {x} right) right) & x ge 1}
$$
(the point being that all arguments of the square roots are now positive).

With some extra work, it is then possible to in fact show that this is indeed $0$ everywhere (with a minor quibble about $x=0$ itself). From here the manipulations are indeed relatively straightforward.