Proving that $\sec\frac\pi{30}=\sqrt{2-\sqrt{5}+\sqrt{15-6\sqrt{5}}}$ I recently saw on this site, the identity
$$\sec\frac\pi{30}=\sqrt{2-\sqrt{5}+\sqrt{15-6\sqrt{5}}}$$
which I instantly wanted to prove.
I know that I can "reduce" the problem to the evaluation of $\cos\frac\pi{15},$ as the rest is easy with the use of the half-angle formula. 
I know that $\cos$ obeys the 'nice' relation
$$\cos nx=T_n(\cos x)$$
where $$T_n(x)=\frac{n}2\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^k}{n-k}{n-k\choose k}(2x)^{n-2k}.$$
Thus, setting $t=\cos\frac\pi{15}$,
$$T_{15}(t)=-1.$$
The only thing left to do is solve for $t$. We can narrow down our search to the values $0<t<1.$
I have never dealt with degree-$15$ polynomials before, so I was hoping one of you could help me out. Thanks!

Staying with your approach (which could be made simpler as said in comments and answers), you have
$$T_{15}(x)=-15 x+560 x^3-6048 x^5+28800 x^7-70400 x^9+92160 x^{11}-61440 x^{13}+16384
   x^{15}$$
but $T_{15}(x)+1$ can be factorized as
$$T_{15}(x)+1=(x+1) (2 x-1)^2 \left(4 x^2-2 x-1\right)^2 \left(16 x^4+8 x^3-16 x^2-8 x+1\right)^2$$ which means that you are left with the quartic
$$16 x^4+8 x^3-16 x^2-8 x+1=0$$ which can be solved with radicals. Its solutions are
$$\left\{x= \frac{1}{8} \left(-1-\sqrt{5}-\sqrt{6
   \left(5-\sqrt{5}\right)}\right)\right\},\left\{x= \frac{1}{8}
   \left(-1-\sqrt{5}+\sqrt{6 \left(5-\sqrt{5}\right)}\right)\right\},\left\{x=
   \frac{1}{8} \left(-1+\sqrt{5}-\sqrt{6
   \left(5+\sqrt{5}\right)}\right)\right\},\left\{x= \frac{1}{8}
   \left(-1+\sqrt{5}+\sqrt{6 \left(5+\sqrt{5}\right)}\right)\right\}$$
