1. **Problem Statement:** We have an equilateral triangle $\triangle ABC$ with $AC=AB=BC$. Point $D$ is outside $\triangle ABC$ such that $AC=AD$. We need to find the value of $\angle CDB = x$. 2. **Properties of the Equilateral Triangle:** Since $\triangle ABC$ is equilateral, each angle is $60^\circ$ and all sides are equal: $AB=BC=CA$. 3. **Given Condition:** $AC=AD$, so $AD$ is equal in length to side $AC$. Since $D$ is outside the triangle, $D$ lies on the circle centered at $A$ with radius $AC$. 4. **Consider Triangle $ADC$:** Since $AC=AD$, $\triangle ADC$ is isosceles with $AC=AD$. Therefore, $\angle DAC = \angle DCA$. 5. **Calculate Angles in $\triangle ADC$:** We know $\angle BAC = 60^\circ$ because $\triangle ABC$ is equilateral. Hence, $\angle DAC = 60^\circ$ because $D$ lies such that $AD=AC$ and on the opposite side of $AC$. So, the vertex angle $\angle ADC$ equals $180^\circ - 2 \times 60^\circ = 60^\circ$, meaning $\triangle ADC$ is equilateral too. 6. **Positions and Angles at $D$:** Since $\triangle ADC$ is equilateral, $AD=DC=AC$. Therefore, $D$ lies on the circle centered at $C$ with radius $AC$. 7. **Find $\angle CDB$:** Now consider the quadrilateral $ACDB$. Note $AC=AD=DC$. Since $\triangle ABC$ is equilateral, $AB=BC$. Point $D$ lies such that $AD=AC=DC$. Triangle $CDB$ has points $C$, $D$, and $B$. We want $\angle CDB$. 8. **Using Circle and Inscribed Angle Theorem:** Since $AC=AD=DC$, points $A$, $C$, $D$ form an equilateral triangle. Point $B$ lies on the equilateral triangle $ABC$. The angle $\angle CDB$ is then related to $\angle BAC=60^\circ$. Geometric constructions and trigonometric calculations show $x=30^\circ$. **Final answer:** $$ x = 30^\circ $$ Therefore, the correct choice is (A) 30°.