equation solving - How to get exact roots of this polynomial?

The equation $$ 64x^7 -112x^5 -8x^4 +56x^3 +8x^2 -7x - 1 = 0 $$ has seven solutions $x = 1$, $x = -\dfrac{1}{2}$ and $x = \cos \dfrac{2n\pi}{11}$, where $n$ runs from $1$ to $5$. With NSolve, I tried

NSolve[64 x^7 - 112 x^5 - 8 x^4 + 56 x^3 + 8 x^2 - 7 x - 1 == 0, x, Reals]

and I get

{{x -> -0.959493}, {x -> -0.654861}, {x -> -0.5}, {x -> -0.142315}, \
{x -> 0.415415}, {x -> 0.841254}, {x -> 1.}}

With Solve, I tried

{{x -> -(1/2)}, {x -> 1}, 
  {x -> Root[1 + 6 #1 - 12 #1^2 - 32 #1^3 + 16 #1^4 + 32 #1^5 &, 1]}, 
  {x -> Root[1 + 6 #1 - 12 #1^2 - 32 #1^3 + 16 #1^4 + 32 #1^5 &, 2]}, 
  {x -> Root[1 + 6 #1 - 12 #1^2 - 32 #1^3 + 16 #1^4 + 32 #1^5 &, 3]}, 
  {x -> Root[1 + 6 #1 - 12 #1^2 - 32 #1^3 + 16 #1^4 + 32 #1^5 &, 4]}, 

  {x -> Root[1 + 6 #1 - 12 #1^2 - 32 #1^3 + 16 #1^4 + 32 #1^5 &, 5]}}

How to get exact solutions of the given equation?

Answer

Since we ask if the numbers $\;x_n = \cos(\frac{2n\pi}{11})\;$ are the actual roots of the polynomial :

p[x_] := 64 x^7 - 112 x^5 - 8 x^4 + 56 x^3 + 8 x^2 - 7 x - 1

any numerical approach cannot be sufficient and in order to prove the statement we should proceed with a symbolic approach. Nevertheless NSolve may guarantee that all the roots could be represented in terms of values of trigonometrical functions like Sin or Cos for real arguments since we have :

And @@ ( -1 <= x <= 1 /. NSolve[ p[x] == 0, x] )

True

The five of the roots are represented in terms of the Root objects, and only two of them have been rewritten by built-in rewrite rules as rational numbers :

r = List @@ Roots[64 x^7 - 112 x^5 - 8 x^4 + 56 x^3 + 8 x^2 - 7 x - 1 == 0, x][[All, 2]];
r[[5 ;;]]

{ Root[1 + 6 #1 - 12 #1^2 - 32 #1^3 + 16 #1^4 + 32 #1^5 &, 5], -(1/2), 1}

Now we can experss r[[;;5]] in terms of Cos (it is possible as we have shown above), we can do it this way :

ArcCos @ r[[;; 5]] // FullSimplify

{ 10π/11, 8π/11, 6 π/11, 4π/11, 2π/11}

We might also use ArcSin as well. Let`s verify if they are equal :

Table[ Root[1 + 6 #1 - 12 #1^2 - 32 #1^3 + 16 #1^4 + 32 #1^5 &, k] - Cos[2(6 - k)π/11],
       {k, 5}] // RootReduce

{0, 0, 0, 0, 0}

Well, indeed these are the roots of the polynomial $p(x)$. One observes that FullSimplify cannot reduce the above Table with the standard built-in rewrite rules unless one uses e.g. Table[ Root[...] - Cos[...], {k,5}]// FullSimplify[#, TransformationFunctions -> RootReduce]&.

Another way which might be helpful in more involved cases would be e.g. mapping PossibleZeroQ, however we have to remember that PossibleZeroQ provides a quick but not always accurate test.

Edit

Since all the roots can be represented as Sin or Cos for real arguments, it would be a good idea to explain what is so specific behind the polynomial p[x]. We can reach a general view working with a transformation pointed out by whuber in the comments.

g[z_] := (p[x] /. x -> (z + 1/z)/2) 2 z^Exponent[p[x], x]
g[z] // Factor

(-1 + z)^2 (1 + z + z^2) (1 + z + z^2 + z^3 + z^4 + z^5 + z^6 + z^7 + z^8 + z^9 + z^10)

All the roots of g[z] are roots of unity :

And @@ RootOfUnityQ[ List @@ Roots[ g[z] == 0, z][[All, 2]] ]

True

moreover all the polynomial factors of g[z] are cyclotomic polynomials, respectively $C^{2}_{1}(z)$, $C_{3}(z)$ and $C_{11}(z)$ (see Cyclotomic), so we have :

Times @@ (Cyclotomic[#, z] & /@ {1, 1, 3, 11}) == g[z] // Factor

True

Following in reverse direction we would generate more polynomials with the roots expressible in terms of $\sin$ or $\cos$ functions on rational multiples of $\pi$.