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If $\cos\frac \pi{n+1}$ is a root of the equation $8x^3+4x^2-4x-1=0$, then find n $(n\in\mathbb N)$

My Attempt:

Let $\theta=\frac\pi{n+1}$, therefore,

$$8\cos^3\theta+4\cos^2\theta-4\cos\theta-1=0$$

Also, $\cos3\theta=4\cos^3\theta-3\cos\theta$ and $\cos2\theta=2\cos^2\theta-1$, so,

$$2(\cos3\theta+3\cos\theta)+2(1+\cos2\theta)-4\cos\theta-1=0\\\implies\cos3\theta+\cos2\theta+\cos\theta=-\frac12\\\implies\frac{\sin\frac{3\theta}2}{\sin\frac\theta2}\cos2\theta=-\frac12\\\implies\frac{\sin\frac{3\pi}{2(n+1)}}{\sin\frac\pi{2(n+1)}}\cos\frac{2\pi}{n+1}=-\frac12$$

How to proceed from here?

EDIT:

Taking hint from Marcos's answer.

$(y+\frac1y)^3+(y+\frac1y)^2-2(y+\frac1y)-1=0$

Adding $1$ to both side,

$(y+\frac1y)^3-1+(y+\frac1y)^2-2(y+\frac1y)+1=1$

$(y+\frac1y-1)((y+\frac1y)^2+(y+\frac1y)+1)+(y+\frac1y-1)^2=1$

$(y+\frac1y-1)((y+\frac1y)^2+2(y+\frac1y))=1$

$(y+\frac1y-1)(y+\frac1y)(y+\frac1y+2)=1$

Not sure what to do with it.

aarbee
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  • @DietrichBurde No, because knowing that $n$ is $6$ gives us different approach from not knowing it. In the linked answer, the starting point is $\cos4\theta=\cos3\theta$. I don't know how to apply that here. – aarbee May 02 '22 at 16:38
  • The roots are shown anyway: "Hence or otherwise, show that the roots of the equation $8x^3+4x^2-4x-1=0$ are $$\cos \frac{2\pi}{7},\cos \frac{4\pi}{7},\cos \frac{6\pi}{7}$$ – Dietrich Burde May 02 '22 at 16:45
  • @DietrichBurde But, there, the angle is visible, so we could have a solid start. In my problem, we don't know the angle, hence the difficulty. That makes it a fresh question, in my opinion. – aarbee May 02 '22 at 16:47
  • @DietrichBurde In the linked answer, they have not solved the equation, they have formed it. – aarbee May 02 '22 at 16:52
  • And they have solved it. The result is $\cos(n\pi/7)$ for $n=2,4,6$. Since a cubic cannot have more than three roots, we are done, also without the angle assumption. – Dietrich Burde May 02 '22 at 16:53
  • @DietrichBurde, there, they are talking about the numerator $n$. Their $7$ in the denominator is fixed. In my problem, we don't have that luxury. We need to find the denominator. – aarbee May 02 '22 at 16:56
  • But since a cubic has only three roots, your possible new root $\cos (\pi/(n+1)$ must coincide with one of these. So you have that luxury for free. – Dietrich Burde May 02 '22 at 16:57
  • @DietrichBurde thankyou so much for trying to help but I am sorry I am not seeing your point. Any chance you could flesh out an answer that might help me to see the connection better? Thanks. – aarbee May 02 '22 at 16:58
  • @DietrichBurde or, can you just tell me what the starting point of my problem shall be? – aarbee May 02 '22 at 17:00
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    @Dietrich Burde.- If $f(x)=8x^3+4x^2-4x-1$ for each of your roots $\cos \frac{2\pi}{7},\cos \frac{4\pi}{7},\cos \frac{6\pi}{7}$ one has $f(x)\ne0$ (verified by Desmos). – Piquito May 03 '22 at 00:04
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    @Piquito No, the roots are correct. I think you must have made a mistake in desmos – Milten May 03 '22 at 06:38
  • @Piquito The duplicate even proves that these are the three roots of $f(x)$. So we are done. – Dietrich Burde May 03 '22 at 08:26
  • @Milten.- If there is a mistake it is not mine but Desmos's. See the COMMENT deleted below, please. – Piquito May 03 '22 at 08:44
  • @Dietrich Burde.- If there is a mistake it is not mine but Desmos's. See the COMMENT deleted below, please (I put it as an answer because I needed the figure). – Piquito May 03 '22 at 08:47
  • @Piquito No problem, it is not your mistake. I just wanted to say that this question has already an answer, because the three roots are well known. This has been shown several times at this site. So this are not "my roots". – Dietrich Burde May 03 '22 at 08:48
  • @Dietrich Burde.- I see that my "your roots" was a show of my deficient English. If you have perceived a kind of irony I apologize. I have a lot of respect for you. – Piquito May 03 '22 at 08:56
  • @Piquito And also, for a CAS the result $10^{-16}$ is like "infinitely small", so likely just zero. The precision is not so high when you compute this. It is only a numerical approximation. – Dietrich Burde May 03 '22 at 09:12
  • See this answer. Note that the equation is not the same. – mathlove Apr 06 '23 at 11:41
  • @mathlove following that link, does that mean in my case, $y^7-1=0?$, so, $y=e^{i(2k\pi)/7}$. So, $2x=2\cos(2k\pi/7)$. But we need $\cos(\frac{\pi}{n+1})$. How do we account for $2k$ here? – aarbee Apr 10 '23 at 11:17
  • I realized that the answer to your question is that there is no such $n$. The roots of your equation are $\cos\frac{2\pi}{7},\cos\frac{4\pi}{7},\cos\frac{6\pi}{7}$. None of these is of the form $\cos\frac{\pi}{n+1}$ where $n\in\mathbb N$. So, I think that it should be $\cos\frac{2\pi}{n+1}$, for example. – mathlove Apr 10 '23 at 17:07
  • @mathlove nice catch, as usual! – aarbee Apr 10 '23 at 18:30

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If you know that a polynomial has a root of the form $\cos\left(\frac{\pi}{n+1}\right)$ then you have to think in roots of unity and cyclotomic polynomials. The idea is to find the right change of variable that transforms your polynomial into a cyclotomic polynomial. Since $8x^3+4x^2-4x-1=(2x)^3+(2x)^2-2(2x)-1$ it seems resonable to look for a change of variable of the form $2x=\dots$ and some standard change can be $2x=y+\frac{1}{y}$. Now, try to make this change of variable and see what you get, you should be able to finish the exercise from here.

Marcos
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