Solve the equation $8x^3-6x+\sqrt2=0$ The solutions of this equation are given. They are $\frac{\sqrt2}{2}$, $\frac{\sqrt6 -\sqrt2}{4}$ and $-\frac{\sqrt6 +\sqrt2}{4}$. However i'm unable to find them on my own. I believe i must make some form of substitution, but i can't find out what to do. Help would be very appreciated. Thanks in advance.

HINT...Let $x=\cos\theta$
The equation becomes $\cos3\theta=-\frac{1}{\sqrt{2}}$
Can you take it from there?
