30 degrees

Problem

Define $X'$ as the unique point on line $AF$ such that $\angle EXF=75^\circ$. In order to prove that $X=X'$ we need to prove that $\angle X'CD=45^\circ$. $\angle X'AC=\angle FAC-\angle CAB=120^\circ-30^\circ=90^\circ$. Note that $\angle AEX'=180^\circ-75^\circ-30^\circ=75^\circ$ so $AX'=AE$. But then note that $\triangle EAC$ is isosceles. So $AX'=AC$ and $\angle \angle ACX'=45^\circ$. Since $\angle ACB=30^\circ$, we see that $\angle X'CD=45^\circ$. So now we can safely deduce that $\angle FXE=75^\circ$. $\blacksquare$

Diagram

Remarks