Solve $y^3-3y-\sqrt{2}=0$ using trigonometry This is a part of a larger question.
I had to show that for $4x^3-3x-\cos 3\alpha=0$ one of the solutions is $\cos \alpha$ and then find the other two solutions. Here they are:
$$4x^3-3x-\cos 3\alpha = (x-\cos \alpha)(2x+\cos \alpha + \sqrt{3} \sin \alpha)(2x+\cos \alpha - \sqrt{3} \sin \alpha)$$
I have to use the above and the following results: $\cos 15^{\circ} = \frac{\sqrt{3}+1}{2\sqrt{2}}$ and $\sin 15^{\circ}=\frac{\sqrt{3}-1}{2\sqrt{2}}$ to find the solutions of the following:
$$y^3-3y-\sqrt{2}=0$$
I assumed that the constant term must be the equivalent of the cosine term and tried to find alpha so that I have one solution and then can derive the other. But since $\arccos {\sqrt{2}}$ is not trivially defined, this is not a correct approach. Or at least it is not correct the way I am doing it. Also I would have to do a bit more trigs since the second polynomial is not equivalent to the first one. There must be an easier, neater solution.
 A: Recall the triplication formula:
$$
\cos3\alpha=4\cos^3\alpha-3\cos\alpha
$$
so, if your equation is $y^3-3y=\sqrt{2}$, you can first set $y=at$, so
$$
a^3t^3-3at=\sqrt{2}
$$
and you'd like that $a^3/3a=4/3$, so you can take $a=2$: $8t^3-6t=\sqrt{2}$; setting $t=\cos\alpha$, we get
$$
\cos3\alpha=\frac{\sqrt{2}}{2}
$$
so
$$
3\alpha=\frac{\pi}{4}+2k\pi
$$
where $k=0$, $k=1$ or $k=2$. Hence
$$
\frac{\pi}{12},\quad\frac{3\pi}{4},\quad\frac{17\pi}{12}
$$
are the solutions for $\alpha$.
Now,
$$
2\cos\frac{\pi}{12}=2\sqrt{\frac{1+\cos(\pi/6)}{2}}=
\sqrt{2+\sqrt{3}}=\frac{\sqrt{6}+\sqrt{2}}{2}
$$
then
$$
2\cos\frac{3\pi}{4}=-\sqrt{2}
$$
and
$$
2\cos\frac{17\pi}{12}=-2\cos\frac{5\pi}{12}=
-2\sqrt{\frac{1+\cos(5\pi/6)}{2}}=
-\sqrt{2-\sqrt{3}}=\frac{\sqrt{6}-\sqrt{2}}{2}
$$
A: Using your hints: If $y$ satisfies $y^3-3y-\sqrt{2}=0$, try  setting $y=kx$ for $k$ to be chosen later. Then $x$ satisfies
$$k^3x^3-3kx-\sqrt2=0\tag1$$
If $k=2$, the LHS of (1) becomes
$$
8x^3-6x-\sqrt2=2\left(4x^3-3x-\frac{\sqrt2}2\right),
$$
which means that $x$ satisfies the famous equation
$$
4x^3-3x-\cos 3\alpha=0\tag{*}
$$
with $\alpha=15^\circ$. Now apply the factorization of (*) you've been given and your two results about $\sin 15^\circ$ and $\cos 15^\circ$ to obtain the three possible solutions for $x$. Finally, get the solutions of the original equation knowing that $y=kx=2x$.
A: If $x=\cos A,$ using $\cos3y=4\cos^3y-3\cos y$
$$\cos3A=\cos3\alpha\implies3A=2n\pi\pm3\alpha$$ where $n$ is any integer
$$\implies A=\dfrac{2n\pi}3+\alpha$$ where $n\equiv0,1,2\pmod3$

Alternatively, $$4x^3-3x-(4\cos^3\alpha-3\cos\alpha)=0$$
$$4(x^3-\cos^3\alpha)-3(x-\cos\alpha)=0$$
Take out $$x-\cos\alpha$$ as common factor & solve the remaining quadratic equation of $x$
A: Thee trigonometric method relies on the trigonometric identity:
$$\cos 3\theta=4\cos^3\theta -3\cos\theta.$$
Set $\;y=A\cos\theta,\enspace A>0$. The equation becomes
$$A^3\cos^3\theta -3A\cos\theta=\sqrt2$$
We choose $A>0$ such that the left-hand side of the equation is proportional to the development of $\cos3\theta $, i.e.
$$\frac{A^3}4=\frac{3A}3\iff A^2=4.$$
Thus we set $y=2\cos\theta $, so that we get the trigonometric equation
$$2\cos3\theta =\sqrt2\iff \cos3\theta =\frac{\sqrt2}2=\cos\frac\pi4,$$
and the solutions are
$$3\theta \equiv\pm\frac\pi4\mod2\pi\iff\theta \equiv\pm\frac\pi{12}\mod\frac{2\pi}3$$
There are $3$ different values for $\cos\theta$, and the different values for $y$ are
$$2\cos\frac\pi{12},\quad2\cos\frac{3\pi}4=-\sqrt2,\quad2\cos\frac{-7\pi}{12}=-\sin\frac\pi{12}$$
Now, the addition formulae yield:
\begin{align*}
\cos\frac\pi{12}&=\cos\Bigl(\frac\pi3-\frac\pi4\Bigr)=\dots=\frac{\sqrt6+\sqrt2}4,\\
\sin\frac\pi{12}&=\sin\Bigl(\frac\pi3-\frac\pi4\Bigr)=\dots=\frac{\sqrt6-\sqrt2}4.
\end{align*}
A: To convert the second equation into the form of the first equation, let $y=2x$ and divide out by $2$ to obtain:
$$4x^3-3x-\frac{\sqrt{2}}{2}=0$$
Thus, the first solution is $\cos 3\alpha = \frac{\sqrt{2}}{2}$ or $\alpha = \pm\frac{\pi}{12}+2n\pi$, $x=\cos{\frac{\pi}{12}}$ or $y=2\cos{\pi/12}$. The other solutions are easily found applying the factorization obtained in the first equation.
