Solving $\cos4\theta=\cos3\theta$ to find the roots of $8x^3+4x^2-4x-1=0$. Find all the solution to the equation
$\cos 4\theta=\cos 3\theta$. Hence or otherwise, show that the roots of the equation
$$8x^3+4x^2-4x-1=0$$ are $$\cos \frac{2\pi}{7},\cos \frac{4\pi}{7},\cos \frac{6\pi}{7}$$
Therefore prove that
$$\sec \frac{2\pi}{7}+\sec \frac{4\pi}{7}+\sec \frac{6\pi}{7}=-4$$
I am slightly stuck! I have solved the first equation using a graphical calculator and found that it is $(2n\pi)/7$ where n is any positive integer. I have then algebraically rewritten $cos(4\theta) = cos(3\theta)$ using double angle identities.
Now I am stuck... Is there are mistake in the question? What would be your model answer?
Thank you so very much.
 A: $\cos(x) = \cos(y) \Leftrightarrow x = 2n\pi \pm y, n\in \Bbb Z$
So,
$4\theta = 2n\pi \pm3\theta  \Rightarrow \begin{cases}\theta = 2n\pi\\\text{ or} \\\theta = \dfrac{2n\pi}{7}\end{cases}$
Now using compound angle formula,
$\begin{align}&\cos4\theta = \cos3\theta \\\Rightarrow &2\cos^2(2\theta )-1 = 4\cos^3\theta-3\cos\theta \\\Rightarrow &2(2\cos^2\theta-1)^2 -1 = 4\cos^3\theta-3\cos\theta \\
\Rightarrow &8\cos^4\theta-8\cos^2\theta+1 = 4\cos^3\theta-3\cos\theta \\
\Rightarrow&8\cos^4\theta-4\cos^3\theta- 8\cos^2\theta+3\cos\theta+1 = 0\end{align}$

We don't need the case where $\theta = 2n\pi$, that'll lead to $\cos\theta =1$.

So, we can factor out $\cos\theta -1 $ from the 4th order equation in $\cos$ (Perform long division or any other suitable method)
$\Rightarrow (\cos\theta-1)(8\cos^3\theta+4\cos^2\theta-4\cos\theta-1) = 0$
$\Rightarrow 8\cos^3\theta+4\cos^2\theta-4\cos\theta-1 =0$ or $\boxed{8x^3+4x^2-4x-1 = 0}$.
As $\cos$ has a periodicity of $2\pi$, we only consider for $n = \pm1,\pm2,\pm3$ or $\cos\frac{\pm2\pi}{7} = \cos\frac{2\pi}{7},\cos\frac{\pm4\pi}{7}=\cos\frac{4\pi}{7},\cos\frac{\pm6\pi}{7}=\cos\frac{6\pi}{7} $.

So the roots are $\cos\frac{2\pi}{7},\cos\frac{4\pi}{7},\cos\frac{6\pi}{7}$


If $a,b,c$ are roots of $Ax^3 +Bx^2+Cx+D = 0$, then
$ab +bc + ca = \frac{C}{A}$ and $abc = -\frac{D}{A} \Rightarrow \frac{ab+bc+ca}{abc}= \frac1a+\frac1b+\frac 1c = -\frac CD$
As $\frac1{\cos x} = \sec x$, here,
$$\frac1a+\frac 1b +\frac 1c =\sec\frac{2\pi}7 +\sec\frac{4\pi}7+\sec\frac{6\pi}7- \frac{-4}{-1} = 4$$
