How find $x$ in a right triangle $ABC$ (${\measuredangle}A=90^\circ$) where ${\measuredangle}DBC={\measuredangle}DCA=x$,${\measuredangle}BAD=5x$? In a right triangle $ABC$ (${\measuredangle}A=90^\circ$) taken in point $D$ such that $BD=AC$, ${\measuredangle}DBC={\measuredangle}DCA=x$,${\measuredangle}BAD=5x$. How find $x$?

W.l.o.g. you may assume the following coordinates:
$$A=(0,0)\quad B=(a,0)\quad C=(0,1)\quad D=(b,c)$$
If you define $d:=\cos(x)$ then you can compute the cosines of multiple angles using a Chebyshev polynomial as
$$\cos(5x)=T_5(d)=16d^5 - 20d^3 + 5d\;.$$
The dot product of two vectors is proportional to the cosine, so you can write the first angle equality as
$$\frac{\langle D-B,C-B\rangle^2}{\lVert D-B\rVert^2\cdot\lVert C-B\rVert^2} = d^2$$
and likewise for the other two angle conditions, the last of which will make use of $\cos^2(5x)$ on its right hand side. The equal length condition is simply expressed as
$$\lVert B-D\rVert^2 = \lVert A-C\rVert^2\;.$$
By using squares in all steps involving the lengths of some vector, we managed to avoid introducing square roots into our equations. So now you have four polynomial equations in four variables, which can be rewritten as
\begin{align*}{}
- a^4 d^2 + a^4 + 2 a^3 b d^2 - 2 a^3 b -  a^2 b^2 d^2 + a^2 b^2 -  a^2 c^2 d^2 \\{}
+ 2 a^2 c -  a^2 d^2 - 2 a b c + 2 a b d^2 -  b^2 d^2 -  c^2 d^2 + c^2 &= 0
\\[2ex]{}
- b^2 d^2 -  c^2 d^2 + c^2 + 2 c d^2 - 2 c -  d^2 + 1 &= 0
\\[2ex]{}
-256 a^2 b^2 d^{10} + 640 a^2 b^2 d^8 - 560 a^2 b^2 d^6 + 200 a^2 b^2 d^4 - 25 a^2 b^2 d^2 \\{}
+ a^2 b^2 - 256 a^2 c^2 d^{10} + 640 a^2 c^2 d^8 - 560 a^2 c^2 d^6 + 200 a^2 c^2 d^4 - 25 a^2 c^2 d^2 &= 0
\\[2ex]{}
a^2 - 2 a b + b^2 + c^2 - 1 &= 0
\end{align*}
Now in theory you could use your favorite computer algebra system to compute solutions for these. Perform some extra checks, like ensuring that the orientations match your expectations, and you can obtain possible $x=\arccos(d)$ from the remaining solutions. In practice, my sage is still busy trying to solve this system. I'll update this answer if it manages to come up with a solution.
While I still don't have a list of all solutions, I can confirm that you have a solution of this system for
$$a=1\qquad b=1-\tfrac12\sqrt3\qquad c=\tfrac12\qquad
d=\tfrac12\sqrt{\sqrt3 + 2}\qquad x=15°$$
as your comment indicated. However, strictly speaking you have $\angle DBC=\angle DCA=-15°$ here, but $\angle BAD=+75°$. So this is a solution except for the orientation of the angles.

