An algebra problem by Akeel Howell

We are given that $x+\dfrac{1}{x}=\sqrt{3}$ . Multiplying by $x$ gives $x^2-\sqrt{3}x+1=0$ .

Applying the quadratic formula, we see that $x = \dfrac{\sqrt{3}\pm i}{2} \\ \implies x = \displaystyle{\left\{\dfrac{\sqrt{3}}{2}+\dfrac{i}{2}, \space \dfrac{\sqrt{3}}{2}-\dfrac{i}{2}\right\}} \\ \therefore x = \large{\space e^{\pi i/6}, \space e^{11\pi i/6}}$

Since an $n^{th}$ root of unity is defined as $\large{e^{2k\pi i/n}}$ for $k = 1,2,3,...,n$ , these are primitive ${(2\times 6)}^{th} = 12^{th}$ roots of unity.