Simplifying complex radicals from trigonometric expression How do I evaluate $ sin(20)$ exactly? [in degrees]
I derived the relationship between $sin(x) $ and $sin(3x)$  where $x = sin(x)$ and $ y = sin(3x)$
http://www.wolframalpha.com/input/?i=-4x%5E3+%2B+3x+%3D+y%2C+solve+for+x
Now I am interested in subsituting $sin(60) $ and moving along but I am not sure which formula will result in me getting a real solution.
http://www.wolframalpha.com/input/?i=-4x%5E3+%2B+3x+%3D+%283%29%5E%281%2F2%29%2F2%2C+solve+for+x
I am curious how to de-nest this mess into something cleaner. I would really like to get rid of all the imaginary numbers but if that is not possible I still feel that this can definitely be de-nested into a simpler looking form even if the radical count does not go down. 
I have tried a couple attempts at substituting stuff back in but my answer seems to change whenever I move thing in and out of the cube root

So the solution to this equation corresponding to the actual value of $\sin(20˚)$ is given by WA as the second solution in your second link as
$$
x = -\frac{1-i \sqrt3}{2\cdot2^{2/3}} \left(-\sqrt{3}+i\right)^{1/3}-\frac14 \left(\frac12 (-\sqrt{3}+i)\right)^{1/3} (1+i \sqrt3) 
$$
We can rewrite this as
$$
x=\left[-\frac{1-i \sqrt3}{2\cdot2^{1/3}} -\frac14  (1+i \sqrt3)\right]\left(\frac12 (-\sqrt{3}+i)\right)^{1/3}
$$
This in turn becomes
$$
x=\left[\left(-\frac14 -\frac1{2 \times2^{1/3} }\right) +
\left(\frac1{2 \times2^{1/3} }-\frac14 \right)i\sqrt3\right]
\left(\frac12 (-\sqrt{3}+i)\right)^{1/3}
$$
Maybe that will be a little easier to work with
