@BalarkaSen @TedShifrin To show that the angle of $30^{\circ}$ is constructible could we do it as followed??
$\cos(3 \theta)=4cos^3(\theta)-3\cos(\theta)$
For $\theta=30^{\circ}$ we have the following:
$\cos90^{\circ}=4\cos^430^{\circ}-3\cos30^{\circ} \Rightarrow 0=4\cos^330^{\circ}-3\cos30^{\circ}$
So, $\cos 30^{\circ}$ satisfies the equation $4x^3-3x=0 \Rightarrow x(4x^2-3)=0$
So, $Irr(\cos 30^{\circ}, \mathbb{Q})=4x^2-3$.
Since $\deg Irr(\cos 30^{\circ }, \mathbb{Q})=2$, $\cos 30^{\circ}$ is constructible.
