Find angle in triangle $ABC$ with cevian line $AD$, such that $AB=CD$.

After asking a middle school math teacher, I got the answer as following.

Make an equilateral triangle $EBC$, and then connect $EA$ and make $DF/\!\!/EC$ which intersects $BE$ at point $F$. Now we have the picture above.

Notice that $∠BCA=∠ECA=30°$, we have $AB=AE$. Then from $∠ABC=40°$, we are able to get that $$∠EAB=180°-2∠AEB=180°-2(60°-∠ABC)=140°$$ From $DF/\!\!/EC$, we could obtain that $EF=CD=AB=AE$, which implies that $$∠EAF=∠EFA=\frac{180°-∠AEB}{2}=80°$$ Then $A,D,B,F$ are in a circle, which implies that $∠DAB=∠DFB=60°$ and the answer is clear.

By your work: $$\sin(x+30^{\circ})=2\sin40^{\circ}\sin{x}$$ or $$\sin{x}\cos30^{\circ}+\cos{x}\sin30^{\circ}=2\sin40^{\circ}\sin{x}$$ or $$\tan{x}=\frac{1}{2(2\sin40^{\circ}-\cos30^{\circ})}.$$ But, $$\frac{1}{2(2\sin40^{\circ}-\cos30^{\circ})}=\frac{1}{2(\sin40^{\circ}+\sin40^{\circ}-\sin60^{\circ})}=$$ $$=\frac{1}{2(\sin40^{\circ}-2\sin10^{\circ}\cos50^{\circ})}=\frac{1}{2\cos50^{\circ}(1-2\sin10^{\circ})}=$$ $$=\frac{\tan50^{\circ}}{2\sin50^{\circ}(1-2\sin10^{\circ})}=\frac{\tan50^{\circ}}{2(\sin50^{\circ}-\cos40^{\circ}+\cos60^{\circ})}=\tan50^{\circ},$$ which says $x=50^{\circ}.$

You can do it also syntheticly.

Draw an equilateral triangle $ABF$ where $F$ is on different side of a line $BC$ then $A$. Let $AE$ cuts $BC$ at $D'$ and prove $D=D'$ i.e. $CD' = AB$.

Note that since $\angle BCA = {1\over 2}\angle BFA $ the point $C$ lies on circle centered at $F$ and $r=FA = FB$. So $FC = AB$. Now easy angle chase we see that $\angle CD'F= \angle = CFD' = 80^{\circ}$ so $CF = CD'$ and you are done.