How to solve $8t^3-4t^2-4t+1=0$ I obtain this cubic equation from solving the trigonometric equation $\sin3x=\sin4x$. I don't know how to solve it to obtain 3 roots.
Thank you!
 A: Added, Saturday: Gauss initiated the study of cyclotomic, umm, numbers, and a simple method for creating polynomials for which they were roots. This is discussed in modern terminology HERE. The jpeg at the bottom is from an 1875 book by Reuschle, with many, many such calculations.  I put a list of such cubics with prime denominator at   http://math.stackexchange.com/questions/2022216/on-the-trigonometric-roots-of-a-cubic/2022887#2022887  With prime denominator, the cubic is $x^3 + x^2 - \left( \frac{p-1}{3} \right)  x + c$ with an integer $c.$ For denominator $9$ it is $x^3 - 3 x + 1$ instead, page 174 in Reuschle.
ORIGINAL. umm. The roots of
$$ x^3 + x^2 - 2 x - 1 $$
are
$$ 2 \cos \frac{2\pi}{7} \; , \; \;  2 \cos \frac{4\pi}{7} \; , \; \; 2 \cos \frac{6\pi}{7} \; . \; \;  $$
Taking $$ x = -2t $$ gives
$$ -8t^3 + 4 t^2 + 4 t - 1 $$

A: 
solving the trigonometric equation $\sin3x=\sin4x$

Hint:  use $\,\sin(a)-\sin(b)=2 \sin\left(\frac{a-b}{2}\right) \cos\left(\frac{a+b}{2}\right)\,$ with $\,a=4x,b=3x\,$.
A: When a cubic equation has  are $3$ real roots, one uses a… trigonometric substitution: $\;t=A\cos \theta$, and choose $A>0$ so we can use the formula for $\cos 3\theta$, and I doubt it is very useful in this case.

Now the trigonometric equation $\;\sin 3x=\sin 4x$ is very simple to
  solve with elementary tools:

Just remember the basics of  trigonometric equations: 
\begin{align}
\sin x&=\sin\alpha\iff \begin{cases}s\equiv\alpha\mod 2\pi\\[-1ex]\qquad\text{ or }\\[-1ex]x\equiv\pi-\alpha\mod 2\pi\end{cases}\\[1ex]
\cos x&=\cos\alpha\iff x \equiv\pm\alpha\mod 2\pi \\[1ex]
\tan x&=\tan\alpha\iff x \equiv\alpha\mod \pi.
\end{align}
Therefore, the solutions of the  trigonometric equation satisfy
\begin{cases}
4x\equiv 3x\mod 2\pi\iff x\equiv 0\mod 2\pi \\[-1ex]\qquad\text{ or }\\[-1ex]
4x\equiv \pi- 3x\mod 2\pi\iff 7x\equiv\pi\mod 2\pi\iff x\equiv\dfrac\pi7\mod \dfrac{2\pi}7.
\end{cases}
A: As Will has said, when you make the substitution $x=2t$, then your polynomial instantly becomes
$$P(x)=x^3+x^2-2x-1$$
The roots can be shown to be equal to$$\begin{align*}x_1 & =2\cos\frac {2\pi}7\\ x_2 & =2\cos\frac {4\pi}7\\x_3 & =2\cos\frac {8\pi}7\end{align*}$$
First, make the transformation $x=u+u^{-1}$ and expand to get that
$$\frac {u^6+u^5+u^4+u^3+u^2+u+1}{u^3}=0$$Multiply both sides by $u-1$ and clear the fraction to get$$u^7=1$$
Through DeMoivre’s Theorem, the solution to $u$ is simply$$u=e^{2k\pi i/7}$$where $k=0,1,2,\ldots,6$. Therefore, by the original substitution$$x=e^{2k\pi i/7}+e^{-2k\pi i/7}=2\cos\frac {2k\pi}7$$
Now, make the appropriate substitutions for $k$ and multiply the result by $2$ to get back to $t$.
A: We have that
$$\sin3x=\sin4x \implies \begin{cases}3x=4x+2k\pi\implies x=2k\pi\\3x=\pi-4x+2k\pi\implies x=\frac{\pi}7+\frac{2k\pi}7\end{cases}$$
