Find $X$ from convex quadrilateral $ABCD$ As title suggests, the question is to find the value of $X$ from the given convex quadrilateral $ABCD$ with some angles given:

My reason for posting this is quite simple, while I did solve this problem (I'll post my solution rest assured), I'm not quite sure if my answer is even correct, let alone the solution being a good approach. So I wanted to see if my answer was correct and what methods there could be to approach this. Any geometric and/or trigonometric approaches are welcome!
 A: So, this'll be my own approach to this problem. I'll add a brief explanation as well:

This is how I go about this:
1.) Locate the circumcenter of $\triangle BDC$ at point $O$ (Note: Yes the circumcenter must lie outside of $\triangle ABC$, this can easily be proven via contradiction, let me know if a discrete proof is needed for this), and connect all three vertices of $\triangle BDC$ to $O$ via segments $OB=OD=OC=AD$. Also join $O$ with $A$ via segment $OA$
2.) Notice that $\triangle OBD$ is in fact equilateral, mark all the appropriate angles, it is also known that $\angle DOC=6x$. Notice that $\triangle BOC$ is an isosceles triangle with $\angle BOC=60+6x$. Further, this implies that $\angle BCO=60-3x$ via the angle sum property.
3.) Notice also that $\triangle BDA$ is also not only isosceles, but congruent to $\triangle BOC$ via the SSS property. This implies that their base angles must also be equivalent. Therefore, we can form an equation for $x$ where $7x=60-3x$. Solving this equation gives us $x=6$ as the final answer.
A: Here is a solution using complex numbers, with the hope you have already studied them.
I will not make any difference between a complex number and the point associated with it in the so-called Argand plane.
WLOG, we can take $B$ as the origin and $A$ as point $1$ (on the real axis).

*

*hypothese $AB=BC$ gives

$$c=e^{i4x}\tag{1}$$

*

*Besides, as $DA=DB$, angle chasing in isosceles triangle $DAB$ gives angle $\hat{D}=\pi-14x$. Therefore, denoting by $R_{\Omega,\theta}$ the rotation with center $\Omega$ and angle $\theta$:

$$R_{D,\pi-14x}(\vec{DB})=\vec{DA} \ \iff \ (0-d)e^{i(\pi-14x)}=a-d \ \iff \ de^{i 14 x }=1-d\tag{2}$$
giving:
$$d=\frac{1}{1-e^{-i 14x}}\tag{3}$$

*

*Finally, exploiting the $\pi/6$ information, we can write:

$$R_{C,\pi/6}(\vec{CD})=r \vec{CB} \ \text{for a certain real number r}$$
giving:
$$(d-c)e^{i \pi/6}=r (0-c) \ \text{for a certain real number r}$$
which amounts to say:
$$r=\frac{c-d}{c}e^{i \pi/6} \ \text{is a real number},\tag{4}$$
a fact that we are going to express by the equality between  $r$ and its conjugate expression:
$$(1-\frac{d}{c})e^{i \pi/6}=(1-\frac{\bar{d}}{\bar{c}})e^{-i \pi/6} \ \iff \ (1-\frac{d}{c})e^{i \pi/3}=(1-\frac{\bar{d}}{\bar{c}}) \tag{5}$$
Now, plugging (1) and (3) into (5) gives an equation with the single unknown $x$.
I leave you the final task to form it.
And then either to solve it or check that $x=\frac{1}{30}\pi$ (the radian measure for $6$ degrees) is a solution.
