2015/057) Three roots of the equation $4p^3 - 3p - 0.5 = 0$ all lie between 1 and -1. Solve

If we let $p = \cos,t$ then we get

$\cos 3t = .5 = \cos \dfrac{\pi}{3}$

so we get $3t$ =$\dfrac{\pi}{3}$ or $\dfrac{7\pi}{3}$ or $\dfrac{13\pi}{3}$

or  $t$ =$\dfrac{\pi}{9}$ or $\dfrac{7\pi}{9}$ or $\dfrac{13\pi}{9}$

or $p$ =  $\cos \dfrac{\pi}{9}$ or $\cos \dfrac{7\pi}{9}$ or $\cos \dfrac{13\pi}{9}$