The traditional solution is to solve the minimum polynomial
$$ \begin{aligned} 64 x^6-112 x^4+56 x^2-7 &= 0\\ \end{aligned} $$
Let us use Cardano's Method, Take $t=4 x^2-\dfrac{7}{3}$, then:
$$ t^3-\frac{7 t}{3}+\frac{7}{27} = 0 $$
which solves t:
$$ t = \sqrt[3]{-\frac{7}{2}+\frac{21 i \sqrt{3}}{2}}+\sqrt[3]{-\frac{7}{2}-\frac{21 i \sqrt{3}}{2} } $$
Choose the appropriate branch, the solution obtained is:
$$ \sin\frac{\pi}{7}=\sqrt{\frac{7}{12}-\frac{1}{24} \left(1-i \sqrt{3}\right) \sqrt[3]{-\frac{7}{2}+\frac{21 i \sqrt{3}}{2}}-\frac{1}{24} \left(1+i \sqrt{3}\right) \sqrt[3]{-\frac{7}{2}-\frac{21 i \sqrt{3}}{2}}} $$
The characteristic is that the quadratic root is set into the cubic root and then the quadratic root is set into.
But I found another version of the solution recorded in a book.
$$ \sin\frac{\pi}{7}=-\frac{\sqrt{7}}{6}+\frac{1}{6}\sqrt[3]{\frac{13\sqrt{7}+3\sqrt{21}i}{2}}+\frac{1}{6}\sqrt[3]{\frac{13\sqrt{7}-3\sqrt{21}i}{2}}\\ $$
This solution is a cubic root of a quadratic root
It is verified that the minimum polynomial is also
$$ 64 x^6-112 x^4+56 x^2-7 $$
How was this kind of solution obtained?
Note that this question is different from Exact values of cos(2π/7) and sin(2π/7), do not merge directly.
That answer also the traditional solution.
