Vtyjkyuk

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!

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 $x^3 + x^2 - left( fracp-13 right) x + c$ 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
$$ 2 cos frac2pi7 ; , ; ; 2 cos frac4pi7 ; , ; ; 2 cos frac6pi7 ; . ; ; $$
Taking $$ x = -2t $$ gives
$$ -8t^3 + 4 t^2 + 4 t - 1 $$

solving the trigonometric equation $sin3x=sin4x$

Hint:  use $,sin(a)-sin(b)=2 sinleft(fraca-b2right) cosleft(fraca+b2right),$ with $,a=4x,b=3x,$.

When a cubic equation has are $3$ real roots, one uses a… trigonometric substitution: $;t=Acos theta$, and choose $A>0$ so we can use the formula for $cos 3theta$, 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:
beginalign
sin x&=sinalphaiff begincasessequivalphamod 2pi\[-1ex]qquadtext or \[-1ex]xequivpi-alphamod 2piendcases\[1ex]
cos x&=cosalphaiff x equivpmalphamod 2pi \[1ex]
tan x&=tanalphaiff x equivalphamod pi.
endalign

Therefore, the solutions of the trigonometric equation satisfy
begincases
4xequiv 3xmod 2piiff xequiv 0mod 2pi \[-1ex]qquadtext or \[-1ex]
4xequiv pi- 3xmod 2piiff 7xequivpimod 2piiff xequivdfracpi7mod dfrac2pi7.
endcases

Solving the equation $8t^3-4t^2-4t+1=0$

If you are familiar with numerical methods, you can use Newton's method to get one numerical solution. After that you might use long division to reduce to a quadratic equation and solve with the pq-formula.

But this will not give 100% accurate solutions.

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$$beginalign*x_1 & =2cosfrac 2pi7\ x_2 & =2cosfrac 4pi7\x_3 & =2cosfrac 8pi7endalign*$$
First, make the transformation $x=u+u^-1$ and expand to get that
$$frac u^6+u^5+u^4+u^3+u^2+u+1u^3=0$$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$$u=e^2kpi i/7$$where $k=0,1,2,ldots,6$. Therefore, by the original substitution$$x=e^2kpi i/7+e^-2kpi i/7=2cosfrac 2kpi7$$
Now, make the appropriate substitutions for $k$ and multiply the result by $2$ to get back to $t$.

We have that

$$sin3x=sin4x implies begincases3x=4x+2kpiimplies x=2kpi\3x=pi-4x+2kpiimplies x=fracpi7+frac2kpi7endcases$$