Solving $x^3 – 3x – \sqrt{2} = 0$ using $x=k\cos\theta$ and $\cos 3\theta = 4 \cos^3 \theta – 3 \cos \theta$

Question

I have proved the first part of this problem, but am stuck.

Prove
$$\cos 3\theta = 4 \cos^3 \theta – 3 \cos \theta$$

Hence solve the equation

$$x^3 – 3x – \sqrt{2} = 0$$

by using a suitable substitution of the form $x = k \cos\theta$, then finding possible values of $\theta$ between $0^\circ$ and $180^\circ$ Give your answers in surd form.

I am working through an A level maths course. I am studying on my own and have looked at the answer in the book which tells me $k = 2$. I don’t know how this was established but can see that using this we can show $\cos 3θ$ = $1/\sqrt{2}$.

But I do not know how to proceed.

Answer 1

If, in the equation $x^3-3x-\sqrt2=0$, you do the substitution $x=2y$, you get $8y^3-6y-\sqrt2=0$, which is equivalent to $4y^3-3y-\frac{\sqrt2}2=0$. But $\frac{\sqrt2}2=\cos\left(\frac\pi4\right)$. So, take $\theta=\frac\pi{12}$ and use the fact that$$4\cos^3\theta-3\cos\theta=\cos(3\theta)=\frac{\sqrt2}2.$$