How would you find the exact roots of $y=x^3+x^2-2x-1$? My friend asked me what the roots of $y=x^3+x^2-2x-1$ was.
I didn't really know and when I graphed it, it had no integer solutions. So I asked him what the answer was, and he said that the $3$ roots were $2\cos\left(\frac {2\pi}{7}\right), 2\cos\left(\frac {4\pi}{7}\right)$ and $2\cos\left(\frac {8\pi}{7}\right)$.

Question: How would you get the roots without using a computer such as Mathematica? Can other equations have roots in Trigonometric forms? 

Anything helps!

Let $p(x) = x^3+x^2-2x-1$, we have $$p(t + t^{-1}) = t^3 + t^2 + t + 1 + t^{-1} + t^{-2} + t^{-3} = \frac{t^7-1}{t^3(t-1)}$$
The RHS has roots of the form $t = e^{\pm \frac{2k\pi}{7}i}$ 
( coming from the $t^7 - 1$ factor in numerator ) 
for $k = 1,2,3$. So $p(x)$ has roots of the form $$e^{\frac{2k\pi}{7} i} + e^{-\frac{2k\pi}{7} i} = 2\cos\left(\frac{2 k\pi}{7}\right)$$ for $k = 1,2,3$.
