is there a solution for equation $\arcsin((1x)^{1/2})=\arccos(x^{1/2})$ in which $x$ is rational number without using $\sin$ and $\cos$ functions in which?

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: $$ \arcsin \left( \sqrt {1x} \right) =i\ln \left( \sqrt {x}+i\sqrt {1x} \right) $$ and $$ \arccos \left( \sqrt {x} \right) =1/2\pi +i\ln \left( \sqrt {1x}+i\sqrt {x} \right) $$ 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 $$\cases{i \left( \ln \left( \sqrt {x}+\sqrt {1x} \right) \ln \left( \sqrt {1x}\sqrt {x} \right) \right) & x\leq 0 \cr 1/2\pi +\arctan \left( {\frac {\sqrt {1x}}{\sqrt {x}}} \right) +\arctan \left( {\frac {\sqrt {x}}{\sqrt {1x}}} \right) & x\in(0,1) \cr i \left( \ln \left( \sqrt {x}\sqrt {1+x} \right) \ln \left( \sqrt {1+x}+\sqrt {x} \right) \right) & x \ge 1} $$ (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. 

