the angle x

We note that

$\begin{aligned} x & = \angle CED = \angle CEA - \angle DEA & \small \blue{\text{Let }\angle CED = \beta \text{ and } \angle DEA = \gamma} \\ \tan x & = \tan (\beta - \gamma) = \dfrac {\tan \beta - \tan \gamma}{1+\tan \beta \tan \gamma} \\ \tan \beta & = \frac {CA}{AE} = \frac {\tan 3\alpha}{\tan \alpha \tan (60^\circ -\alpha)} = \frac {3t-t^3}{1-3t^2} \cdot \frac {1+\sqrt 3t}{t(\sqrt 3-t)} & \small \blue{\text{Let }AB=1 \text{ and }t = \tan \alpha} \\ & = \frac {t(\sqrt 3-t)(\sqrt 3+t)}{(1-\sqrt 3t)(1+\sqrt 3t)} \cdot \frac {1+\sqrt 3t}{t(\sqrt 3-t)} \\ & = \frac {\sqrt 3+t}{1-\sqrt 3t} = \tan (60^\circ + \alpha) \\ \implies \beta & = 60^\circ + \alpha \\ \tan \gamma & = \frac {DA}{AE} = \frac 1{\tan (60^\circ - \alpha)} = \cot (60^\circ - \alpha) = \tan (30^\circ + \alpha) \\ \gamma & = 30^\circ + \alpha \\ \implies x & = \beta - \gamma = \boxed{30}^\circ \end{aligned}$

I'll post my solution shortly...