Submitted by Josh on Sun, 10/29/2017 - 17:49

I'm on part 3 but i'm confused what i am meant to do now. I have got 8x^3 -6x-root2=0 after subbing in y as 2x but i'm not sure how I can relate this to the previous question because in ii the equation involves cos3theta but in this equation we have root 2.

Any help would be appreciated.

## Divide by 2

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.