Finding minimum polynomial of $e^{i\pi/6 }$
Let $\zeta:=e^{i\pi/6}$. As you note, $\zeta$ is a root of $t^4-t^2+1$, because it is clearly a root of $$t^6+1=(t^2+1)(t^4-t^2+1),$$ and it is not hard to check that it is not a root of $t^2+1$. For the same reason, also $\zeta^5$, $\zeta^7$ and $\zeta^{11}$ must be roots of $t^4-t^2+1$. So apparently $$t^4-t^2+1=(t-\zeta)(t-\zeta^5)(t-\zeta^7)(t-\zeta^{11}).$$ If $t^4-t^2+1$ were reducible, then either $t-\zeta^m\in\Bbb{Q}[t]$ for some $m\in\{1,5,7,11\}$, which is clearly false, or otherwse $(t-\zeta)(t-\zeta^m)\in\Bbb{Q}[t]$ for distinct $m\in\{5,7,11\}$. Trying all three options shows that this is not the case, so $t^4-t^2+1$ is irreducible over $\Bbb{Q}$.
Alternatively, by the rational root test it is clear that $t^4-t^2+1$ has no roots in $\Bbb{Q}$. We can check by brute force that it is not a product of two quadratics by solving $$(t^2+at+b)(t^2+ct+d)=t^4-t^2+1,$$ for $a,b,c,d\in\Bbb{Q}$, which comes down to solving the system of equations \begin{eqnarray*} a+c&=&0,\\ ac+b+d&=&-1,\\ ad+bc&=&0,\\ bd&=&1, \end{eqnarray*} First it is immediate that $c=-a$ and $d=b^{-1}$, so the third equation becomes $$ab^{-1}-ba=0\qquad\text{ and so }\qquad ab^2=a,$$ that is, either $a=c=0$ or $b=d=\pm1$. In the first case the second equation becomes $$b+b^{-1}=-1\qquad\text{ and so }\qquad b^2-b+1=0,$$ which is impossible. In the second case the second equation becomes $$-a^2\pm2=-1,$$ which is also impossible. So there is no such factorization, so $t^4-t^2+1$ is irreducible.
Here's a slick way of showing irreducibility of $x^4- x^2+1$, using uniqueness of factorization in both $\Bbb Q[x]$ and $K[x]$, where $K$ is the field of Gaussian numbers, $\Bbb Q(i)$.
Note that $x^4-x^2+1=(x^2+ix-1)(x^2-ix-1)$, and that this is a factorization into irreducibles over $K$. If the original quartic had any factorization over $\Bbb Q$, it would also be a factorization over $K$, but this product of quadratics is the only $K$-factorization that there is. So the original quartic is irreducible over the rationals.