Thursday, Novemeber 10, 2016: I found the method of Gauss written up in Galois Theory by David Cox, chapter 9, section 2. The method lends itself to computer programming, otherwise your eyes start to blur after a few of the calculations. I posted an answer with polynomials extending the primes up to 311. Gauss, clever guy.
To save space, here are just the polynomials for primes $p \equiv 1 \pmod {10}$ up to $1000.$
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jagy@phobeusjunior:~$ ./quintic_cyclic_gauss_loop | grep exps
x^5 + x^4 - 4 x^3 - 3 x^2 + 3 x + 1 p 11 p.root 2 exps 10^k d = 11^4
x^5 + x^4 - 12 x^3 - 21 x^2 + 1 x + 5 p 31 p.root 3 exps 6^k d = 5^2 31^4
x^5 + x^4 - 16 x^3 + 5 x^2 + 21 x - 9 p 41 p.root 6 exps 3^k d = 3^6 41^4
x^5 + x^4 - 24 x^3 - 17 x^2 + 41 x - 13 p 61 p.root 2 exps 21^k d = 29^2 61^4
x^5 + x^4 - 28 x^3 + 37 x^2 + 25 x + 1 p 71 p.root 7 exps 23^k d = 23^2 71^4
x^5 + x^4 - 40 x^3 + 93 x^2 - 21 x - 17 p 101 p.root 2 exps 32^k d = 17^2 101^4
x^5 + x^4 - 52 x^3 - 89 x^2 + 109 x + 193 p 131 p.root 2 exps 18^k d = 79^2 131^4
x^5 + x^4 - 60 x^3 - 12 x^2 + 784 x + 128 p 151 p.root 6 exps 23^k d = 2^18 151^4
x^5 + x^4 - 72 x^3 - 123 x^2 + 223 x - 49 p 181 p.root 2 exps 17^k d = 7^2 149^2 181^4
x^5 + x^4 - 76 x^3 - 359 x^2 - 437 x - 155 p 191 p.root 19 exps 11^k d = 5^2 11^2 191^4
x^5 + x^4 - 84 x^3 - 59 x^2 + 1661 x + 269 p 211 p.root 2 exps 26^k d = 31^2 67^2 211^4
x^5 + x^4 - 96 x^3 - 212 x^2 + 1232 x + 512 p 241 p.root 7 exps 11^k d = 2^16 11^2 241^4
x^5 + x^4 - 100 x^3 - 20 x^2 + 1504 x + 1024 p 251 p.root 6 exps 2^k d = 2^18 5^4 251^4
x^5 + x^4 - 108 x^3 - 401 x^2 - 13 x + 845 p 271 p.root 6 exps 12^k d = 5^2 13^4 271^4
x^5 + x^4 - 112 x^3 - 191 x^2 + 2257 x + 967 p 281 p.root 3 exps 6^k d = 193^2 281^4
x^5 + x^4 - 124 x^3 + 535 x^2 - 413 x - 539 p 311 p.root 17 exps 11^k d = 7^4 13^2 311^4
x^5 + x^4 - 132 x^3 - 887 x^2 - 1843 x - 1027 p 331 p.root 3 exps 13^k d = 13^2 31^2 331^4
x^5 + x^4 - 160 x^3 + 369 x^2 + 879 x - 29 p 401 p.root 3 exps 26^k d = 29^2 401^4 433^2
x^5 + x^4 - 168 x^3 + 219 x^2 + 3853 x - 3517 p 421 p.root 2 exps 32^k d = 223^2 239^2 421^4
x^5 + x^4 - 172 x^3 - 724 x^2 + 1824 x + 1728 p 431 p.root 7 exps 47^k d = 2^20 3^4 431^4
x^5 + x^4 - 184 x^3 - 129 x^2 + 4551 x + 5419 p 461 p.root 2 exps 13^k d = 163^2 461^4 491^2
x^5 + x^4 - 196 x^3 + 59 x^2 + 2019 x + 1377 p 491 p.root 2 exps 32^k d = 3^4 17^2 229^2 491^4
x^5 + x^4 - 208 x^3 - 771 x^2 + 4143 x + 2083 p 521 p.root 3 exps 24^k d = 61^2 521^4 577^2
x^5 + x^4 - 216 x^3 + 1147 x^2 - 805 x - 2629 p 541 p.root 2 exps 11^k d = 11^2 311^2 541^4
x^5 + x^4 - 228 x^3 + 868 x^2 + 3056 x - 7552 p 571 p.root 3 exps 2^k d = 2^22 31^2 571^4
x^5 + x^4 - 240 x^3 + 1755 x^2 - 3731 x + 2399 p 601 p.root 7 exps 17^k d = 5^2 13^2 17^2 601^4
x^5 + x^4 - 252 x^3 + 2095 x^2 - 5785 x + 5069 p 631 p.root 3 exps 24^k d = 89^2 631^4
x^5 + x^4 - 256 x^3 - 564 x^2 + 5328 x - 5120 p 641 p.root 3 exps 21^k d = 2^16 5^2 61^2 641^4
x^5 + x^4 - 264 x^3 - 185 x^2 + 16837 x + 4851 p 661 p.root 2 exps 32^k d = 3^16 7^2 661^4
x^5 + x^4 - 276 x^3 - 1299 x^2 + 5329 x + 15581 p 691 p.root 3 exps 11^k d = 379^2 397^2 691^4
x^5 + x^4 - 280 x^3 + 2047 x^2 - 3791 x + 1699 p 701 p.root 2 exps 23^k d = 17^2 19^2 23^2 701^4
x^5 + x^4 - 300 x^3 - 2313 x^2 - 3761 x - 571 p 751 p.root 3 exps 11^k d = 41^2 631^2 751^4
x^5 + x^4 - 304 x^3 + 2831 x^2 - 8925 x + 8775 p 761 p.root 6 exps 3^k d = 3^4 5^2 23^2 761^4
x^5 + x^4 - 324 x^3 - 3471 x^2 - 12431 x - 13603 p 811 p.root 3 exps 12^k d = 7^4 47^2 811^4
x^5 + x^4 - 328 x^3 - 1215 x^2 + 3573 x + 2179 p 821 p.root 2 exps 32^k d = 37^4 109^2 821^4
x^5 + x^4 - 352 x^3 - 2361 x^2 + 4257 x + 9967 p 881 p.root 3 exps 29^k d = 29^2 881^4 953^2
x^5 + x^4 - 364 x^3 - 2988 x^2 - 1392 x + 9856 p 911 p.root 17 exps 22^k d = 2^18 7^2 11^2 911^4
x^5 + x^4 - 376 x^3 + 3877 x^2 - 13445 x + 15271 p 941 p.root 2 exps 12^k d = 191^2 941^4
x^5 + x^4 - 388 x^3 + 1476 x^2 + 8304 x + 7168 p 971 p.root 6 exps 2^k d = 2^20 7^2 13^2 971^4
x^5 + x^4 - 396 x^3 + 2101 x^2 + 8039 x - 1819 p 991 p.root 6 exps 30^k d = 107^2 991^4 1399^2
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ORIGINAL QUESTION: I should emphasize that I am interested in quintics with five real irrational roots, Galois group cyclic $\mathbb Z_5,$ such that the roots can be expressed as sums of roots of unity (in conjugate pairs), therefore as sums of cosines of rational multiples of $\pi.$
The wording at the wikipedia section on solvable cyclic quintics suggests that there is an infinite sequence of such examples as they display. However, they give no source for this sub-section. As you will see below, i had no trouble extending their recipe for primes $101$ and $131,$ but the items i found for prime $151$ have the wrong Galois groups. I have jumped to prime $181$ and bigger $p = 10 n + 1,$ we will see what happens. I have stuck with their recipe…note that, by a strict integer translation, it is reasonable to also consider $x^5 + 2 x^4 + more,$ or $x^5 + 3 x^4 + more,$ or $x^5 + 4 x^4 + more.$ There is plenty of literature on $x^5 + e x^3 + stuff.$ Oh, with $p = 10 n + 1$ prime, our polynomial is
$$ x^5 + x^4 – 4 n x^3 + a x^2 + b x + c $$ i have been assuming that we want the discriminant to be a square, in particular $w^2 p^4,$ where $w$ is not divisible by $p.$
The questions are: where did wikipedia find this material, also are there more.
https://en.wikipedia.org/wiki/Quintic_function#Other_solvable_quintics
How to solve a cyclic quintic in radicals?
Solve this tough fifth degree equation.
$$ x^5 + x^4 – 4 x^3 – 3 x^2 + 3 x + 1 $$
$$ \Delta = 11^4 $$
$$ $$
$$ x^5 + x^4 – 12 x^3 – 21 x^2 + x + 5 $$
$$ \Delta = 5^2 \; 31^4 $$
$$ $$
$$ x^5 + x^4 – 16 x^3 + 5 x^2 + 21 x – 9 $$
$$ \Delta = 3^6 \; 41^4 $$
$$ $$
$$ x^5 + x^4 – 24 x^3 – 17 x^2 + 41 x – 13 $$
$$ \Delta = 29^2 \; 61^4 $$
$$ $$
$$ x^5 + x^4 – 28 x^3 + 37 x^2 + 25 x + 1 $$
$$ \Delta = 23^2 \; 71^4 $$
$$ $$
$$ x^5 + x^4 – 40 x^3 + 93 x^2 – 21 x – 17 $$
$$ \Delta = 17^2 \; 101^4 $$
$$ $$
$$ x^5 + x^4 – 52 x^3 – 89 x^2 + 109 x + 193 $$
$$ \Delta = 79^2 \; 131^4 $$
$$ $$
Tito(151)
$$ x^5 + x^4 -60 x^3 -12 x^2 + 784 x + 128 $$
$$ \Delta = 2^{18} \; 151^4 $$
$$ $$
$$ x^5 + x^4 -72 x^3 -123 x^2 + 223 x -49 $$
$$ \Delta = 7^2 \; 149^2 \; 181^4 $$
$$ $$
Emma Lehmer(191)
$$ x^5 + x^4 – 76 x^3 -359 x^2 – 437 x – 155 $$
$$ \Delta = 5^2 \; 11^2 \; 191^4 $$
$$ $$
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Best Answer
Yes, there are infinitely many cyclic quintics as parameterized by the Emma Lehmer quintic
$$F(y)=y^5 + n^2y^4 - (2n^3 + 6n^2 + 10n + 10)y^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)y^2 + (n^3 + 4n^2 + 10n + 10)y + 1 = 0$$
This also obeys
$$ y_1 y_2 + y_2 y_3 + y_3 y_4 + y_4 y_5 + y_5 y_1 - (y_1 y_3 + y_3 y_5 + y_5 y_2 + y_2 y_4 + y_4 y_1) = 0$$
Let $p=25 + 25 n + 15 n^2 + 5 n^3 + n^4$. Then the discriminant of $F(y)$ is
$$D = (7 + 10 n + 5 n^2 + n^3)^2\,p^4$$
Also, note that if $m=n+1$, then $n\,p=m^5 + 5m^3 + 5m - 11$. A root is given by $$y = a+b\sum_{k=1}^{(p-1)/5}\,{\zeta_p}^{c^k}$$
with root of unity $\zeta_p = e^{2\pi i/p},\,$ for some integer $a,b,c$. See this MO post for the formulas for $a,b,c$.
P.S. While $p=151$ does not belong to this family, I find that,
$$x^5 + x^4 - 60x^3 - 12x^2 + 784x + 128 = 0$$
with discriminant $d=2^{18}151^4$ has the root $\displaystyle x=\sum_{k=1}^{30}e^{2\pi\, i\, c^k/151}$ for $c=23$. The Mathematica command to find these quintics is,
for prime $p\equiv1\pmod{10}$. Inspecting the resulting table of candidate quintics, identical ones with small coefficients will stand out and which gives the correct choice of $c$.