Your solution looks correct to me.
I'm going to write two solutions.
Solution 1 :
Let us take $E$ such that $\triangle{ACE}$ is an equilateral triangle. ($E$ is on the same side of $AC$ as $D$.)
Now, let us first show that $B,D,E$ are collinear.
Since $AB=AC=AE$, we see that $\triangle{ABE}$ is an isosceles triangle, and so we get $$\angle{ABE}=\frac{1}{2}(180^\circ-\angle{BAE})=\frac{1}{2}(180^\circ-100^\circ)=40^\circ=\angle{ABD}$$
Therefore, it follows that $B,D,E$ are collinear.
Since $\triangle{DCA}$ is congruent to $\triangle{DCE}$, we obtain
$$x=\angle{DAC}=\angle{DEC}=\color{red}{20^\circ}$$
Solution 2 :
This is a trigonometric approach.
Using the law of sines in $\triangle{ABD}$, we have
$$AD=\frac{\sin 40^\circ}{\sin(100^\circ-x)}AB\tag1$$
Using the law of sines in $\triangle{ADC}$, we have
$$AD=\frac{\sin 30^\circ}{\sin(150^\circ-x)}AC\tag2$$
From $(1)(2)$ with $AB=AC$, we get
$$\frac{\sin 40^\circ}{\sin(100^\circ-x)}=\frac{\sin 30^\circ}{\sin(150^\circ-x)}$$
i.e.
$$\sin 40^\circ\ \bigg(\frac 12\cos x+\frac{\sqrt 3}{2}\sin x\bigg)=\frac 12(\sin 100^\circ\cos x-\cos 100^\circ\sin x)$$
Dividing the both sides by $\frac 12\cos x$, and solving it for $\tan x$ give $$\begin{align}\tan x&=\frac{\sin 100^\circ-\sin 40^\circ}{\sqrt 3\sin 40^\circ+\cos 100^\circ}
\\\\&=\frac{\sin(90^\circ +10^\circ)-\sin(30^\circ+10^\circ)}{\sqrt 3\sin(30^\circ+10^\circ)+\cos(90^\circ +10^\circ)}
\\\\&=\frac{\frac 12\cos 10^\circ-\frac{\sqrt 3}{2}\sin 10^\circ}{\frac{\sqrt 3}{2}\cos 10^\circ+\frac 12\sin 10^\circ}
\\\\&=\frac{\frac{1}{\sqrt 3}-\tan 10^\circ}{1+\frac{1}{\sqrt 3}\tan 10^\circ}
\\\\&=\frac{\tan 30^\circ -\tan 10^\circ}{1+\tan 30^\circ \tan 10^\circ}
\\\\&=\tan (30^\circ -10^\circ)
\\\\&=\tan 20^\circ\end{align}$$
Therefore, we get $x=\color{red}{20^\circ}$.
This is going to be my approach to the problem:
Here's my approach for the problem:
1.) Extend $CD$ to meet $E$ that lies on the circumference. Join the center of the circle $O$ with $E$ via segment $OE$. Since $OE$ is radius, we know that $OE=9$. Notice that $\angle ECB$ is the inscribed angle of $\angle EOC$, therefore we can conclude that $\angle EOB=2\alpha$
2.) Notice that $\angle EOA=\angle EDB=180-2\alpha$, therefore $OE=ED=9$. Now we can use the intersecting chords theorem (which can easily be proven via similarity of triangles in any general cyclic quadrilateral) and conclude that:
$$9\cdot CD=6\cdot12=72$$
$$\Rightarrow CD=8$$
Best Answer
This is my approach:
Note that $\angle AED=5x$. This means that $\angle ACD=3x$. Therefore, $AD=CD$. Let $F$ be a point on $BD$ such that $\angle BFC=\angle BAC=4x$. This implies that $\angle FCD=\angle FDC=2x$, $FD=FC$ and $\angle FCA=x$. This proves that quadrilateral $ABCF$ is cyclic, therefore $\angle FAC=x$ as well. This implies that $FD=FC=FA$, therefore $\angle FDA=2x$ as well.
Hence, $10x=180^\circ$, $x=18^\circ$