# How do I get the solutions of a cubic equation in trigonometric form?

I know $x^3-3 x+1=0$ has three roots that can be expressed in trigonometric form: $\{2\sin(10^\circ),\,-2\cos(20^\circ),\,2\cos(40^\circ)\}$.

How can I get this result with Mathematica?

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Use e.g. ComplexExpand@(x /. Solve[x^3 - 3 x + 1 == 0, x]). This question concerns the same issue as mathematica.stackexchange.com/questions/17269/… –  Artes Jan 25 '13 at 15:31
Another related question : mathematica.stackexchange.com/questions/14726/… –  Artes Jan 25 '13 at 15:43
Thanks. But I want Cos[π/9] + Sqrt[3] Sin[π/9] can be simplify to 2 Cos[(2 π)/9] –  mathe Jan 25 '13 at 15:58