Calculate the value of angle $ACB$

Denoting $\angle ACB=\gamma$ (the one we need to find), you have that $$CD=2BD+AD\Leftrightarrow \\ \frac{CD}{BD}=2+\frac{AD}{BD}\quad (1) $$ Apply sine law in $\triangle ADB$ and $\triangle CDB$: $$\frac{CD}{BD}=\frac{\sin7\gamma}{\sin\gamma}\\ \frac{AD}{BD}=\frac{\sin4\gamma}{\sin2\gamma}=2\cos2\gamma $$ so let $\sin\gamma=x$ and substitute in $(1)$. $\cos2\gamma=1-2x^2$ and $\sin7\gamma=7x-56x^3+112x^5-64x^7$ (see here). You get the equation $$64x^6-112x^4+52x^2-3=0\Leftrightarrow \\ (4x^2-3)(16x^4-16x^2+1)=0 $$ which is solvable by letting $t=x^2$. And you have to take into account that $7\gamma<180^{\circ}$, so $0<x<\sin\frac{180^{\circ}}{7}<\sin\frac{180^{\circ}}{6}=\frac 12\Rightarrow 0<t<\frac 14$. We get $t=\frac{2-\sqrt{3}}{4}\Rightarrow x=\frac{\sqrt{2-\sqrt{3}}}{2} \Rightarrow \gamma=15^{\circ}$ (see here for a table of trig. values).