$4-$ $[$ Trigonometric Equations $]$

Solve the equations
a) $x^{3}-3 x-1=0$

b) $x^{3}-3 x-\sqrt{3}=0$.

Explicitly state all roots of these equations.
By the Vieta's method (see problem $\underline{61275}$), making the substitution $x=2 t$, we get the equations $4 t^{3}-3 x-1 / 2=0$ and

$4 t^{3}-3 t-\frac{\sqrt{3}}{2}=0 . \quad \arccos \frac{1}{2}=\frac{\pi}{3}, \quad \arccos \frac{\sqrt{3}}{2}=\frac{\pi}{6}$, so (by the formulas from problem 61275) the roots of the first equation are

$t_{1}=\cos \pi / 9, t_{2}=\cos 7 \pi / 9, t_{3}=\cos 13 \pi / 9$, and of the second $t_{1}=\cos \pi / 18, t_{2}=\cos 13 \pi / 18, t_{3}=\cos 25 \pi / 18$.