You have used $y=2x$ and got $8x^3-6x-\sqrt{2}=0$. If you then divide by 2 this becomes $4x^3 - 3x -\frac{\sqrt{2}}2=0$. If we let $\cos 3 \alpha = \frac{\sqrt{2}}2$ then this is the equation in part (ii). Solving $\cos 3 \alpha = \frac{\sqrt{2}}2$ gives $3\alpha = 45^{\circ}$, i.e. $\alpha = 15^{\circ}$ (which ties in with part (i)!). Let me know if you want any further pointers.