In[64]:= Solve[x^3 - 3 x - 1 == 0, x, Reals]

Out[64]= {{x -> Root[-1 - 3 #1 + #1^3 &, 1]}, {x ->  Root[-1 - 3 #1 + #1^3 &, 2]}, {x -> Root[-1 - 3 #1 + #1^3 &, 3]}}

In[67]:= N[{{x -> Root[-1 - 3 #1 + #1^3 &, 1]}, {x -> Root[-1 - 3 #1 + #1^3 &, 2]}, {x -> Root[-1 - 3 #1 + #1^3 &, 3]}}]

Out[67]= {{x -> -1.53209}, {x -> -0.347296}, {x -> 1.87939}}
In[68]:= {{x -> -1.532088886237956`}, {x -> -0.3472963553338607`}, {x \-> 1.8793852415718169`}}

ExpToTrig[ToRadicals[Root[-1 - 3 #1 + #1^3 &, 1]]] // Re

Thank you for pointing out my misunderstanding. I saw Cubics false in the details and assumed that was the default value.

In[1]:= FullSimplify[x /. Solve[x^3 - 3 x - 1 == 0, x]]

Out[1]= {2 Cos[\[Pi]/9], Root[-1 - 3 #1 + #1^3 &, 1], -Cos[\[Pi]/9] + Sqrt[3] Sin[\[Pi]/9]}

In[2]:= ExpToTrig[ToRadicals[Root[-1 - 3 #1 + #1^3 &, 1]]]

Out[2]= -(1/2) Cos[\[Pi]/9] + 1/2 I Sqrt[3] Cos[\[Pi]/9] - Cos[(2 \[Pi])/9] - 1/2 I Sin[\[Pi]/9] - 1/2 Sqrt[3] Sin[\[Pi]/9] - I Sin[(2 \[Pi])/9]

FullSimplify only the complex terms

In[3]:= -(1/2) Cos[\[Pi]/9] + FullSimplify[1/2 I Sqrt[3] Cos[\[Pi]/9] - 1/2 I Sin[\[Pi]/9] - I Sin[(2 \[Pi])/9]] - Cos[(2 \[Pi])/9] - 1/2 Sqrt[3] Sin[\[Pi]/9]

Out[3]= -(1/2) Cos[\[Pi]/9] - Cos[(2 \[Pi])/9] - 1/2 Sqrt[3] Sin[\[Pi]/9]

In the help page for Solve click on Details and Options and read all the extra documentation for Solve.

In that find that including the option Cubics->True will give you what you want

Solve[x^3 - 3 x - 1 == 0, x, Cubics -> True]

Using FullSimplify on the result will translate two out of the three into trig form.

Using ExpToTrig and FullSimplify in various ways on parts of the third one can get it into trig form.