Using De Moivre's to find roots of a polynomial

a) Use De Moivre's theorem to express $\frac{\sin 8\theta}{\sin\theta \cos\theta}$ as a polynomial in$s$, where $s=\sin\theta$

b) Hence solve the equation $x^6-6x^4+10x^2-4=0$

I've been able to do the first question and I worked out the answer to be $8(1-10s^2+24s^4-16s^6)$ but I can't seem to be able to do part b) because I can't see a relationship between the two polynomials.

Any help is appreciated thanks :)

Hint: $\;4\cdot(1-10s^2+24s^4-16s^6) = -\big((2s)^6 - 6 \cdot (2s)^4 + 10 \cdot (2s)^2 - 4\big)\,$.
The key here is to recognise that the equation $x^{6}-6x^{4}+10x^{2}-4=0$ can be changed into $$4(1-10s^{2}+24s^{4}-16s^{6})=0$$
via the substitution $x=2\sin\theta$. Then you can use the identity you proved in the first part to find the roots of the equation in $x$.
$$x^6-6x^4+10x^2-4=(x^2-2)(x^4-4x^2+2)=(x^2-2)[(x^2-2)^2-2]=0$$
Thus, let $x^2-2=0$ or $x^4-4x^2+2=0$. We obtain $$x=\pm\sqrt{2},$$or$$x=\pm\sqrt{2\pm\sqrt{2}}.$$