Solving for the roots of a solvable quintic
In the Wikipedia page about quintics, there was a list of quintics that could be solved with trigonometric roots.
For example:$$x^5+x^4-4x^3-3x^2+3x+1\tag1$$ has roots of the form $2\cos \frac {2k\pi}{11}$ $$x^5+x^4-16x^3+5x^2+21x-9=0\tag2$$ has roots of the form $\sum_{k=0}^{7}e^{\frac {2\pi i 3^k}{41}}$
Question: Is there a formula or underlying structure to find the roots of the quintics?
Wikipedia says that the roots are the sums of the first $n$-th roots of unity with $n=10k+1$, so I'm guessing we first have to find $n$, or do we first have to find $k$?
Any help is appreciated.
ADDED: I remembered where I had seen this before, with lower degree; let $\alpha \neq 1$ be a seventh root of unity, $$ \alpha^7 = 1, $$ and take $$ \gamma = \alpha + \alpha^6, $$ $$ \gamma^2 = \alpha^2 + 2 + \alpha^5, $$ $$ \gamma^3 = \alpha^3 + 3 \alpha + 3 \alpha^6 + \alpha^4. $$ Therefore $$ \gamma^3 + \gamma^2 - 2 \gamma - 1 = \alpha^3 + \alpha^2 + \alpha + 1 + \alpha^6 + \alpha^5 + \alpha^4 = 0 $$ This is a cubic with three irrational roots, and is https://en.wikipedia.org/wiki/Casus_irreducibilis
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I simply checked the one that seems easiest. Cute. Let $\omega \neq 1$ be an 11th root of unity, $$ \omega^{11} = 1, $$ so that a root is $$ \beta = \omega + \omega^{10}$$ $$ \beta^2 = \omega^2 + 2 + \omega^9, $$ $$ \beta^3 = \omega^3 + 3 \omega + 3 \omega^{10} + \omega^8, $$ $$ \beta^4 = \omega^4 + 4 \omega^2 + 6 + 4 \omega^9 + \omega^7, $$ $$ \beta^5 = \omega^5 + 5 \omega^3 + 10 \omega + 10 \omega^{10} + 5 \omega^8 + \omega^6. $$ Then $$ 1 + 3 \beta - 3 \beta^2 - 4 \beta^3 + \omega^4 + \omega^5 = 1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 + \omega^7 + \omega^8 + \omega^9 + \omega^{10} = \frac{\omega^{11} - 1}{\omega - 1} = 0 $$
The quintic above and a few more are listed in Wiki Quintic. I cannot tell the source of these, how the list may be extended, or whether this family gives all such possibilities. I checked with gp-pari, for each prime $p = 10n+1,$ the discriminant of their $x^5 + x^4 - 4n x^3 + a x^2 + b x + c$ was $\Delta = w^2 p^4,$ with $w \neq 0 \pmod p.$ For the quintic above, $p = 11,$ and $w = 1.$
Here is an early post that discusses things, How to solve a cyclic quintic in radicals?
Easy enough to find cubic monics of the form $x^3 + x^2 + b x + c$ with no rational roots and square discriminant, so that the Galois group is $\mathbb Z_3.$ Not that there is no loss in having either that form or $x^3 + bx + c,$ as a pure (integer) translation changes the coefficient of $x^2$ by $3,$ and the differences between $x^3 + x^2 + b x + c$ and $x^3 - x^2 + b x - c$ are not important for this problem.
sqrt d b c d sqrt d
7 -142 -701 49 7
7 -2 -1 49 7
13 -4 1 169 13
19 -6 -7 361 19
37 -12 11 1369 37
49 -16 -29 2401 49
56 -9 -1 3136 56
62 -10 -8 3844 62
65 -30 53 4225 65
65 -56 -181 4225 65
79 -26 41 6241 79
86 -14 8 7396 86
91 -16 13 8281 91
91 -44 -127 8281 91
91 -86 -337 8281 91
97 -32 -79 9409 97
104 -17 -25 10816 104
133 -82 259 17689 133
139 -46 103 19321 139
152 -25 31 23104 152
163 -54 -169 26569 163
172 -100 352 29584 172
182 -30 -64 33124 182
183 -20 -9 33489 183
201 -22 5 40401 201
203 -30 41 41209 203
209 -44 -121 43681 209
217 -72 209 47089 217
219 -24 -27 47961 219
247 -82 -311 61009 247
248 -72 -256 61504 248
254 -42 80 64516 254
273 -30 27 74529 273
287 -30 -43 82369 287
296 -49 -137 87616 296
301 -44 83 90601 301
309 -34 -61 95481 309
313 -104 371 97969 313
325 -30 -25 105625 325
349 -116 -517 121801 349
392 -65 167 153664 392
399 -44 69 159201 399
427 -142 601 182329 427
436 -36 -4 190096 436
446 -74 -256 198916 446
448 -37 -29 200704 448
453 -50 -123 205209 453
469 -156 -799 219961 469
496 -41 23 246016 496
520 -121 -545 270400 520
532 -44 -64 283024 532
566 -94 304 320356 566
579 -64 143 335241 579
581 -114 419 337561 581
589 -44 -7 346921 589
611 -82 235 373321 611
628 -52 64 394384 628
632 -105 -433 399424 632
651 -72 -225 423801 651
679 -86 251 461041 679
688 -57 -121 473344 688
728 -65 -169 529984 728
776 -129 503 602176 776
813 -90 261 660969 813
832 -69 131 692224 832
845 -56 -25 714025 845
851 -86 233 724201 851
sqrt d b c d sqrt d
I am not sure if there is any simple formula for the root structure of polynomials having a solvable polynomial. The only thing is that the subnormal series tells you the order in which you adjoin radicals. It does not tell you which number in each field formed you take a radical of to make the next field, but if you make the correct choices, then the final field is the splitting field, and the root of the polynomial will be just linear combinations over the base field, of the elements of the splitting field.
For example: $1\triangleleft G_2\triangleleft V\triangleleft A_4\triangleleft S_4$, where $G_2$ is any order two group containing a double-transposition has the following.
$\text{ord}(S_4/A_4)=2$
$\text{ord}(A_4/V)=3$
$\text{ord}(V/G_2)=2$
$\text{ord}(G_2/1)=2$
Thus the general quartic's splitting field is constructed by these steps:
Step 1: Pick a number in the base field and adjoin it's square root.
Step 2: Pick a number in the new field and adjoin it's cube root and $e^{2\pi i/3}$.
Step 3: Pick a number in the new field and adjoin it's square root.
Step 4: Pick a number in the new field and adjoin it's square root, thus producing the splitting field.
By the way, for your two quintics, we know that: $$e^{\pm 2\pi i/5}=\frac{-1+\sqrt{5}\pm i\sqrt{10+2\sqrt{5}}}{4}$$ $$e^{\pm 4\pi i/5}=\frac{-1-\sqrt{5}\pm i\sqrt{10-2\sqrt{5}}}{4}$$
For $x^5+x^4-4x^3-3x^2+3x+1=0$:
If $(a,b,k)$ is any of $(-4,4,2)$, $(2,-4,4)$, $(4,0,6)$, $(-2,-2,8)$, $(0,2,10)$ then \begin{align*} 2\cos \frac{2k\pi}{11}=-\frac{1}{5} &-\frac{1}{10}e^{a\pi i/5}\sqrt[5]{7832-2200\sqrt{5}+(1320+880\sqrt{5})i\sqrt{10+2\sqrt{5}}}\\ &-\frac{1}{10}e^{-a\pi i/5}\sqrt[5]{7832-2200\sqrt{5}-(1320+880\sqrt{5})i\sqrt{10+2\sqrt{5}}}\\ &-\frac{1}{10}e^{b\pi i/5}\sqrt[5]{7832+2200\sqrt{5}+(1320-880\sqrt{5})i\sqrt{10-2\sqrt{5}}}\\ &-\frac{1}{10}e^{-b\pi i/5}\sqrt[5]{7832+2200\sqrt{5}-(1320-880\sqrt{5})i\sqrt{10-2\sqrt{5}}} \end{align*}
For $x^5+x^4-16x^3+5x^2+21x-9=0$:
If $(a,b,n)$ is any of $(0,-2,1)$, $(-4,0,2)$, $(2,2,4)$, $(-2,4,8)$, $(4,-4,16)$ then
\begin{align*} \sum_{k=0}^7e^{2n\pi i3^k/41}=-\frac{1}{5} &+\frac{1}{10}e^{a\pi i/5}\sqrt[5]{321768-8200\sqrt{5}+(9840+14760\sqrt{5})i\sqrt{10+2\sqrt{5}}}\\ &+\frac{1}{10}e^{-a\pi i/5}\sqrt[5]{321768-8200\sqrt{5}-(9840+14760\sqrt{5})i\sqrt{10+2\sqrt{5}}}\\ &+\frac{1}{10}e^{b\pi i/5}\sqrt[5]{321768+8200\sqrt{5}+(9840-14760\sqrt{5})i\sqrt{10-2\sqrt{5}}}\\ &+\frac{1}{10}e^{-b\pi i/5}\sqrt[5]{321768+8200\sqrt{5}-(9840-14760\sqrt{5})i\sqrt{10-2\sqrt{5}}} \end{align*}