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 $x$?
 A: 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.

A: This is NOT a solution. It is just some of the interesting findings in the course of seeking the required solution.
I have to give that up because I cannot think of any method of incorporating the given 5x into my work. 
I just hope that someone can have an elegant geometric solution and some of my findings maybe useful.



A: this is geo method to prove that only one solution for this question.
Assume we already know the solution of $ x=15°$
then if $\measuredangle B< 45°$ ,see picture below:

note $\angle BDC= \angle DCA$, so $AC$ tangent $BDC$ circumcircle $K$ ,and $BD=AC$ which means $D$ is intersect point of two circles.(ie, orange and blue)
$M$ is midpoint of $BC, A'M \perp BC ,(A',L)=A'M \cap $circle $MC.KC\perp AC, K=KC\cap A'M,L,K$is the circumcenter of $BD'C$ and $BDC.  D' $ is the solution of $A'BC$ so it is also the intersect point point of two circles( green and brown).since $\measuredangle C >45° \implies LM>KM \implies \stackrel{\frown}{BDC}$ out $\stackrel{\frown}{ BD'C} ,BD'>BD \implies  \measuredangle DBC>\measuredangle D'BC =15°. $ 
edit2:
to prove $\angle BAD< \angle BA'D'$ it is hard to explain by geometry so we set parameter $p=\dfrac{1}{\tan{C}}\le 1,MC=1.MB=-1 \implies k_{AD}=\dfrac{p(1+2p^2-\sqrt{p^4+p^2+1})}{1-p^2\sqrt{p^4+p^2+1}}$
$ k_{AD}$ can be postive, but in this case $\measuredangle BAD < \measuredangle BA'D'$. so we only care $ k_{AD} <0$. ie. $p^2\sqrt{p^4+p^2+1}>1$, in this case, $ k_{AD}$ is mono increasing function of $p$. 
to make easy, we prove $f(p)=\dfrac{(1+2p^2-\sqrt{p^4+p^2+1})}{p^2\sqrt{p^4+p^2+1}-1}$ is mono decreasing function. 
$x=p^2, f(p)=f(x)=\dfrac{(1+2x-\sqrt{x^2+x+1})}{x\sqrt{x^2+x+1}-1}=\dfrac{\dfrac{1}{x}-\sqrt{\dfrac{1}{x^2}+\dfrac{1}{x}+1}+2}{\sqrt{x^2+x+1}-\dfrac{1}{x}}$ 
it is trivial $\sqrt{x^2+x+1}-\dfrac{1}{x} >0$ is mono increasing function. 
$g(z)=z-\sqrt{z^2+z+1}, g'(z)=1-\dfrac{z+\dfrac{1}{2}}{\sqrt{z^2+z+1}}>0 \implies g(z)$ is mono increasing.$\implies g(\dfrac{1}{x})$ is mono decreasing function.$\implies f(p)$ is mono decreasing.
so   $ k_{AD}$ get max when $p=1 \implies \measuredangle BAD \le \measuredangle BA'D'$
(end edit2)
we done. 
when $\measuredangle B>45°$ see picture below:

it is same way that we prove $\measuredangle DBC <15° $ and $\measuredangle BAD>75°$
A: As suggested by the OP, if we can show that the line through D parallel to AB is the perpendicular bisector of AC, then x = … = 15 degrees. The following shows how this can be achieved.
Constructions:-
(1) Draw the rectangular coordinate system X’DX and Y’DY.
(2) Draw a circle using D as center, radius = DB. Then AC = DB = DY = DY’.
(3) Join AB and also join YB such that YB cuts DX at P (as shown in the first figure below.)

Move on to the second figure.

From $\measuredangle Y’DP + \measuredangle PBY’ = 90 + 90 = 180$, Y’DPB is a cyclic quadrilateral. [opp. angles supp.]
Then, r = q = p [angles in in the same segment + base angle isosceles triangle]
Therefore, PY = PY’.
Move on to the third figure.
By properties of isosceles triangle, m = n.

Right shift AC along AB  to A'D'C' where A' is the intersection of AB and Y'P; (similar for C').
The light yellow triangle is congruent to the dark yellow triangle. [ASA]
Therefore, A'D' = D'C'.
Because, A'D'C' is just the image of AKC; where K is the intersection of AC and X'X, AK = KC.
Thus, X'X is the perpendicular bisector of AC.
The required result then follows. (See OP's comment.) 
