Points $A, B, C$ and $D$ are on a circle of diameter $1$, and $X$ is on diameter $\overline{AD}$. If $BX=CX$ and $3 \angle BAC=\angle BXC=36^{\circ}$, then $AX=$
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draw(Circle((0,0),10));
draw((-10,0)--(8,6)--(2,0)--(8,-6)--cycle);
draw((-10,0)--(10,0));
dot((-10,0));
dot((2,0));
dot((10,0));
dot((8,6));
dot((8,-6));
label("A", (-10,0), W);
label("B", (8,6), NE);
label("C", (8,-6), SE);
label("D", (10,0), E);
label("X", (2,0), NW);
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$ \textbf{(A)}\ \cos 6^{\circ}\cos 12^{\circ} \sec 18^{\circ} \qquad\textbf{(B)}\ \cos 6^{\circ}\sin 12^{\circ} \csc 18^{\circ} \qquad\textbf{(C)}\ \cos 6^{\circ}\sin 12^{\circ} \sec 18^{\circ} \\ \qquad\textbf{(D)}\ \sin 6^{\circ}\sin 12^{\circ} \csc 18^{\circ} \qquad\textbf{(E)}\ \sin 6^{\circ} \sin 12^{\circ} \sec 18^{\circ} $
1. **Mark the center of the circle \( O \) and draw the radius \( \overline{OB} \).** 
   - Since \( A, B, C, D \) are on a circle of diameter 1, the radius \( OA = OB = OC = OD = \frac{1}{2} \).

2. **Focus on the upper triangle \( ABX \).**
   - Given \( \angle BXC = 36^\circ \), we can deduce \( \angle BXD = 18^\circ \) because \( X \) is on the diameter \( \overline{AD} \).

3. **Perform angle chasing:**
   - Since \( \angle BAC = 12^\circ \) (because \( 3 \angle BAC = 36^\circ \)), we have:
     - \( \angle BAO = 6^\circ \)
     - \( \angle ABO = 6^\circ \)
     - \( \angle ABX = 12^\circ \)
     - \( \angle BOX = 12^\circ \)

4. **Use the Law of Sines on triangle \( ABX \):**
   \[
   \frac{\sin(\angle ABX)}{AX} = \frac{\sin(\angle AXB)}{AB}
   \]
   - Given \( \angle ABX = 12^\circ \) and \( \angle AXB = 18^\circ \) (since \( \angle AXB = \angle BXD \)), we get:
   \[
   AX = \frac{AB \sin 12^\circ}{\sin 18^\circ}
   \]

5. **Use the Law of Sines on triangle \( AOB \):**
   \[
   \frac{\sin(\angle ABO)}{AO} = \frac{\sin(\angle AOB)}{AB}
   \]
   - Given \( \angle ABO = 6^\circ \) and \( AO = \frac{1}{2} \), we get:
   \[
   AB = \frac{\frac{1}{2} \sin 12^\circ}{\sin 6^\circ}
   \]
   - Simplify using the double-angle identity \( \sin 12^\circ = 2 \sin 6^\circ \cos 6^\circ \):
   \[
   AB = \frac{\frac{1}{2} \cdot 2 \sin 6^\circ \cos 6^\circ}{\sin 6^\circ} = \cos 6^\circ
   \]

6. **Substitute \( AB = \cos 6^\circ \) into the expression for \( AX \):**
   \[
   AX = \frac{\cos 6^\circ \sin 12^\circ}{\sin 18^\circ}
   \]

7. **Simplify the expression:**
   - Using the identity \( \sin 18^\circ = \frac{\sqrt{5} - 1}{4} \) and \( \csc 18^\circ = \frac{4}{\sqrt{5} - 1} \):
   \[
   AX = \cos 6^\circ \sin 12^\circ \csc 18^\circ
   \]

The final answer is \(\boxed{\cos 6^\circ \sin 12^\circ \csc 18^\circ}\)