In the book https://ogeometrie-cip.blogspot.com/2021/02/getting-to-know-new-problem-book.html
see PROBLEM 28, at page 115.
I have not found a solution to this problem.
I had to look at the solution given in the book. On page 120 it says just that...
(Long after that I found out about the author's solution: see below, in the contribution of Silviu Boga)
I posted the problem on several forums, let's see what consequences it will have...
Posted on Facebook, 2/19/2021 https://www.facebook.com/petre.ciobanu.5621/posts/708103799874909
Posted on AOPS, 2/20/2021 https://artofproblemsolving.com/community/c4h2458976_equation_for_sinpi14
SOLUTION CiP
Answer CiP
The roots of the polynomial are $sin\frac{\pi}{14},\;\;-sin\frac{3\pi}{14}\;\;and\;\;sin\frac{5\pi}{14}$.
Or, complementarily expressed $cos\frac{\pi}{7},\;\;-cos\frac{2\pi}{7}\;\;and\;\;cos\frac{3\pi}{7}$.
Solution
According to de Moivre's formula
$(cos \alpha +\imath \cdot sin \alpha)^{7}=cos 7 \alpha+\imath \cdot sin 7\alpha,\;(cos \alpha-\imath \cdot sin \alpha)^{7}=cos 7 \alpha-\imath \cdot sin 7\alpha$
express $sin7\alpha=\frac{(cos\alpha+\imath \cdot sin \alpha)^{7}-(cos\alpha-\imath \cdot sin\alpha)^{7}}{2\imath}$. We develop $(cos \alpha \pm \imath \cdot sin\alpha)^{7}$ with Newton's binomial theorem , considering that $\imath ^{2}=\imath ^{6}=-1,\;\imath ^{3}=\imath ^{7}=-\imath,$
$\imath ^{4}=1,\;\imath ^{5}=\imath$, and we obtain $\sin7\alpha=$
$=\frac{1}{2\imath}\cdot [cos^{7}\alpha+7\imath\; cos^{6}\alpha sin \alpha-21 cos^{5}\alpha sin^{2}\alpha-35\imath\; cos^{4}\alpha sin^{3} \alpha +35cos ^{3} \alpha sin^{4} \alpha +21 \imath\; cos^{2} \alpha sin^{5} \alpha -7 cos \alpha sin^{6} \alpha-$
$-\imath \; sin^{7} \alpha -(cos^{7}\alpha-7\imath\; cos^{6}\alpha sin \alpha -21 cos^{5}\alpha sin^{2}\alpha-35 \imath\; cos^{4}\alpha sin^{3}\alpha +35 cos^{3}\alpha sin^{4} \alpha -21 \imath \;cos^{2}\alpha sin^{5} \alpha -$
$-7 cos \alpha sin^{6} \alpha +\imath \; sin^{7} \alpha )]$.
At the result $sin 7\alpha=7cos^{6} sin \alpha -35 cos^{4} \alpha sin^{3} \alpha +21 cos^{2}\alpha sin^{5} \alpha -sin^{7} \alpha$
we replace further $cos^{2}\alpha=1-sin^{2}\alpha, \;\;cos^{4} \alpha =1-2sin^{2}\alpha +sin^{4}\alpha,\;$
$\;cos^{6}\alpha=1-3sin^{2}\alpha +3sin^{4}\alpha-sin^{6} \alpha$, and after other calculations we obtain
(1) $sin 7\alpha =-64sin^{7}\alpha+112sin^{5}\alpha-56sin^{3}\alpha +7sin \alpha$.
(See Bromwich's formulas (24) in Weisstein, Eric W. "Multiple-Angle Formulas.")
We put in the formula (1) $\alpha=\frac{\pi}{14}$. Since $sin 7\cdot \frac{\pi}{14}=sin \frac{\pi}{2}=1$, we obtain that $x=sin \frac{\pi}{14}$ is the root of the seventh degree equation
(2) $ 64 \cdot x^{7}-112 \cdot x^{5}+56\cdot x^{3}-7\cdot x+1=0$.
IF WE ARE LUCKY ENOUGH, we can factorize the polynomial on the left, and find the form equivalent to the equation (2)
$(x+1)\cdot (8\cdot x^{3}-4\cdot x^{2} -4 \cdot x+1)^{2}=0$.
From the above line, it is clear to whom $x=sin\frac{\pi}{14}$ is the root.
$\blacksquare$
..................................................................................................................................
REMARK 1
We write formula (1) in the form
(3) $sin7\alpha-1=-(sin\alpha+1)(8sin^{3}\alpha-4sin^{2}\alpha-4sin\alpha+1)^{2}$.
The values for which $sin7\alpha-1=0$ are $\alpha_{k}=\frac{\pi}{14}+k\cdot \frac{2\pi}{7},\;\;k \in \mathbb{Z}$. So
(4) $(sin \alpha_{k}+1)(8sin^{3}\alpha_{k}-4sin^{2}\alpha_{k}-4sin \alpha_{k}+1)^{2}=0$.
We analyze only cases $k\in \left \{0,\;1,...,6\;\right \}$ because $\alpha_{k+7m}\equiv \alpha_{k}\;\;(mod\;2\pi)$.
The value $\alpha_{5}=\frac{\pi}{14}+\frac{10\pi}{7}=\frac{21\pi}{14}=\frac{3\pi}{2}$ corresponds to the case $sin\alpha_{5}=-1$, which makes zero the first factor in the relation (4). Further,
$sin\alpha_{0}=sin\alpha_{3}$, because $\alpha_{0}+\alpha_{3}=\frac{\pi}{14}+\frac{13\pi}{14}=\pi$,
$sin\alpha_{1}=sin\alpha_{2}$, because $\alpha_{1}+\alpha_{2}=\frac{5\pi}{14}+\frac{9\pi}{14}=\pi$,
$sin\alpha_{4}=sin\alpha_{6}$, because $\alpha_{4}+\alpha_{6}=\frac{17\pi}{14}+\frac{25\pi}{14}=3\pi$.
So the given polynomial $8\cdot x^{3}-4\cdot x^{2}-4\cdot x+1$ has three different roots $sin\alpha_{0}=\sin\frac{\pi}{14},\;sin\alpha_{1}=sin\frac{5\pi}{14},\;sin\alpha_{4}=sin\frac{17\pi}{14}=sin(\pi+\frac{3\pi}{14})=-sin\frac{3\pi}{14}$. We get the first answer.
Observing that $\frac{\pi}{2}-\alpha_{0}=\frac{3\pi}{7},\;\frac{\pi}{2}-\alpha_{1}=\frac{\pi}{7},\;\frac{\pi}{2}-\frac{3\pi}{14}=\frac{2\pi}{7}$, we can still express the answer $x_{1}=cos\frac{\pi}{7},\;x_{2}=-cos\frac{2\pi}{7},\;x_{3}=cos\frac{3\pi}{7}$.
(End Rem #1)
REMARK 2
Writing, for the three roots found, some of Viete's relationships, we have identities
(5) $sin\frac{\pi}{14}-sin\frac{3\pi}{14}+sin\frac{5\pi}{14}=\frac{1}{2}$,
(6) $cos\frac{\pi}{7}-cos\frac{2\pi}{7}+cos\frac{3\pi}{7}=\frac{1}{2}$,
(7) $sin\frac{\pi}{14}\cdot sin\frac{3\pi}{14}\cdot sin\frac{5\pi}{14}=\frac{1}{8}$,
(8) $cos\frac{\pi}{7}\cdot cos\frac{2\pi}{7}\cdot cso\frac{3\pi}{7}=\frac{1}{8}$.
(End Rem #2)
$\bigstar$ $\bigstar$ $\bigstar$
P.S. Here ENDS the solitary adventure, starting from the attempt to solve the problem in the statement. I followed the path suggested by the indication in the book.With the price of a substantial number of scribbled sheets of paper, I am still glad that I solved more than was required:I determined ALL the roots of the given equation.
$\bigstar$ $\bigstar$ $\bigstar$
During the days I worked on this problem, I was stimulated by the help of "fellow mathematicians" who sent me ideas or even other methods to solve this problem. I mention them below, trying not to miss any. My thanks to everyone.
Feb 20, 2021, 6:23 PM AOPS- the user vanstraelen suggested the development of $sin7\alpha$ in terms of $sin\alpha$. He gets the formula (1) which he writes in the form (3).
(It all happened 65 minutes after I posted the question. Others here did not answe.)
Now, about those on Facebook.
Andreea Schier [19, Feb at 23:03] put up a photo
I kind of twisted my neck, then thought about rotating it
She chooses, as was most normal, to calculate the value of the polynomial $8x^{3}-4x^{2}-4x+1$ for $x=sin\frac{\pi}{14}$. It goes without saying that he uses the formula $8\cdot sin^{3}\alpha =6\cdot sin\alpha -2 \cdot sin3\alpha$ (obtained from $sin 3\alpha=3\cdot sin \alpha-4 \cdot sin^{3} \alpha$). However, it mentions the formula $sin^{2} \alpha=\frac{1-cos 2 \alpha}{2}$ as well as a transition from the cosine to the sine of the complementary angle. An expression is reached in the sinuses of various multiples of $\frac{\pi}{14}$, but only at the first power:
$S=2\cdot sin\frac{\pi}{14}-2\cdot sin\frac{3\pi}{14}+2\cdot sin \frac{5\pi}{14}-1$.
She proves that $S=0$, calculating for this $S \cdot sin\frac{\pi}{7}$ ( it transforms sine products into sums with the formula $2\cdot sin \alpha \cdot sin \beta=cos(\alpha-\beta)-cos(\alpha+\beta)$ and all terms are canceled).
I notice that, by the same calculation as hers, a demonstration of the Viete type relationship (5) is obtained.
$\bigstar$
In her comment Marcelina Popa starts from the equation $sin 4\alpha=sin 3\alpha$. This equation has in the open interval $(\;0,\;\pi)$ only the solutions $\alpha_{k}=(2k+1)\cdot \frac{\pi}{7}\;\;k=0,\;1,\;2$.
Equivalently, the equation is written successively (remember that in the interval $(\;0,\;\pi)$)
$4\cdot sin\alpha \cdot cos\alpha\cdot cos2\alpha=sin\alpha\cdot (3-4sin^{2}\alpha)$
$\overset{sin\alpha \neq 0}{\Leftrightarrow} 4\cdot cos \alpha\cdot (2cos^{2}\alpha-1)=3-4(1-cos^{2}\alpha)$
$\Leftrightarrow \;8\cdot cos^{3}\alpha-4\cdot cos^{2}\alpha-4\cdot cos \alpha+1=0$.
As a result, the equation $8x^{3}-4x^{2}-4x+1=0$ is verified for $cos\frac{\pi}{7},\;cos\frac{3\pi}{7}=sin\frac{\pi}{14},\;cos\frac{5\pi}{7}$.
$\bigstar$
Another calculation-based answer gave Dumitru Pietreanu. He starts from equality
$sin(\frac{3\pi}{14}+\frac{4\pi}{14})=sin\frac{\pi}{2}$,
which he develops using known formulas, finding that $x=sin\frac{\pi}{14}$ is the root of a polynomial $Q(x)$ and this can be expressed as $Q(x)=P^{2}(x)\cdot (x+1)$ where $P(x)$ is the given polynomial.
Indeed, after the first step $sin\frac{3\pi}{14}\cdot cos\frac{4\pi}{14}+cos\frac{3\pi}{14}\cdot sin\frac{4\pi}{14}=1$. If $x=sin\frac{\pi}{14}$, then $cos^{2}\frac{\pi}{14}=1-x^{2},\;sin\frac{2\pi}{14}=2x\cdot cos\frac{\pi}{14},\;cos\frac{2\pi}{14}=1-2x^{2},$
$\;sin\frac{3\pi}{14}=3x-4x^{3},\;cos\frac{3\pi}{14}=cos\frac{\pi}{14}\cdot (4cos^{2}\frac{\pi}{14}-3)=cos\frac{\pi}{14}\cdot [4(1-x^{2})-3]=...,$
$sin\frac{4\pi}{14}=4x\cdot cos \frac{\pi}{14}\cdot cos\frac{2\pi}{14}=...,\;cos\frac{4\pi}{14}=2cos^{2}\frac{2\pi}{14}-1=...$.
In the end it results $-64x^{7}+112x^{5}-56x^{3}+7x=1$, so just the formula (2).
He also congratulated Marcel Tena for his contribution.
$\bigstar$
The one who found the mentioned book from other sources, also identified the solutions (two) of the author, on page 296, is
The first solution exploits the fact that $cos \frac{4\pi}{7}+\imath \cdot sin \frac{4\pi}{7} \neq 1$ verifies the equation $z^{7}=1$ (they are all $cos (k\cdot \frac{2\pi}{7})+\imath \cdot sin (k\cdot\frac{2\pi}{7}), \;k=0,\;1,...,6$. !!! There was a small mistake, probably transcription, where I circled in the picture: it should be written "u" instead of "1".
The second solution is similar to ours.
$\bigstar$
Ghimisi Dumitrel used the complex number $\zeta=cos\frac{\pi}{14}+\imath\cdot sin\frac{\pi}{14}$. We have $\zeta^{7}=\imath$. But $\zeta^{7}-\imath=\zeta^{7}+\imath^{7}=(\zeta+\imath)(\zeta^{6}-\zeta^{5}\imath-\zeta^{4}+\zeta^{3}\imath+\zeta^{2}-\zeta\imath -1)$
$\Rightarrow \;\zeta^{6}-\zeta^{5}\imath-\zeta^{4}+\zeta^{3}\imath+\zeta^{2}-\zeta\imath -1=0$.
With $sin\frac{\pi}{14}=\frac{\zeta-\bar{\zeta}}{2}=\frac{\zeta^{2}-1}{2\imath z}$ is calculated $8sin^{3}\frac{\pi}{14}-4sin^{2}\frac{\pi}{14}-4sin\frac{\pi}{14}+1=$
$=8\cdot (\frac{\zeta^{2}-1}{2\imath z})^{3}-4\cdot (\frac{\zeta^{2}-1}{2\imath z})^{2}-4\cdot \frac{\zeta^{2}-1}{2\imath z}+1=-\frac{1}{\imath \zeta^{3}}\cdot [(\zeta^{2}-1)^{3}-\imath \zeta (\zeta^{2}-1)^{2}+2\zeta^{2}(\zeta^{2}-1)-\imath \zeta^{3}]=$
$=-\frac{1}{\imath \zeta^{3}}\cdot [\zeta^{6}-\imath \zeta^{5}-\zeta^{4}+\imath \zeta^{3}+\zeta^{2}-\imath \zeta-1]=0$.
$\bigstar$
Doru GHERASA presented two methods of solving.
In the first solution, he observes that $sin\frac{\pi}{14}=cos(\frac{\pi}{2}-\frac{\pi}{14})=cos\frac{6\pi}{14}$, so $\frac{\pi}{14}$ is a solution of the equation
(9) $sin \alpha =cos 6\alpha$.
Equation (9) is successively transformed into $sin \alpha = 4cos^{3}2\alpha-3cos2\alpha $
$\Leftrightarrow\;sin\alpha=4(1-2sin^{2}\alpha)^{3}-3(1-2sin^{2}\alpha)\;\Leftrightarrow x=4(1-2x^{2})^{3}-3(1-2x^{2})$, so $sin\frac{\pi}{14}$ is a solution of equation
(10) $32x^{6}-48x^{4}+18x^{2}+x-1=0$.
He notices the factorization of (10)
$(8x^{3}-4x^{2}-4x+1)(x+1)(4x^{2}-2x-1)=0$
and, since $0<sin\frac{\pi}{14}<\frac{1}{2}$, it can be the root only for the first factor.
In the second method, we start from de Moivre's formula
$(a+\imath b)^{7}\overset{def}{=}(cos\frac{\pi}{14}+\imath sin \frac{\pi}{14})^{7}=cos\frac{\pi}{2}+\imath sin\frac{\pi}{2}=\;\imath$.
The imaginary part of the equality, developing on the left side with Newton's binomial theorem, involves
$7a^{6}b-35a^{4}b^{3}+21a^{2}b^{5}-b^{7}=1\;\;\;\overset{a^{2}=1-b^{2}}{\Rightarrow}\;\;\; 64b^{7}-112b^{5}+56b^{3}-7b+1=0$, similar to (2). Then mention that the last equation can be written
$(8b^{3}-4b^{2}-4b+1)^{2}(b+1)=0.$
$\bigstar$
AT LAST, BUT NOT LEST, we thanks Mr Marcel ȚENA. I reported the problem to him for the first time. He also had that book. He is, among many other things, the author of the book "RADACINILE UNITATII".
In a discussion with him, I asked his opinion on which cyclotomic polynomial between $\Phi_{14}$ or $\Phi_{7}$ to use. From his answers I realized that it is possible with both. The solution now has only a few lines:
The cyclotomic polynomial $\Phi_{14}$ has, by definition, one of its roots
$\zeta=cos(3\cdot \frac{2\pi}{14})+\imath \cdot sin(3\cdot \frac{2\pi}{14})=cos\frac{3\pi}{7}+\imath \cdot sin\frac{3\pi}{7}$.
But $\Phi_{14}(X)=\Phi_{7}( -X)$ (Theorem 16, page 24) and 7 being prime number we have $\Phi_{7}(X)=X^{6}+X^{5}+X^{4}+X^{3}+X^{2}+X+1$ (Theorem 11, page 20). So $\zeta$ is a root of polynomial $X^{6}-X^{5}+X^{4}-X^{3}+X^{2}-X+1$.
So $\zeta$ verifies the equation
$\zeta ^{6}-\zeta ^{5}+\zeta ^{4}-\zeta ^{3}+\zeta ^{2}-\zeta+1=0$
$\Leftrightarrow \;\;(\zeta ^{3}+\frac{1}{\zeta ^{3}})-(\zeta ^{2}+\frac{1}{\zeta ^{2}})+(\zeta+\frac{1}{\zeta})-1=0$.
We replace above $\zeta +\frac{1}{\zeta}=2cos\frac{3\pi}{7}=sin(\frac{\pi}{2}-\frac{3\pi}{7})=sin \frac{\pi}{14}=2x$,
$\zeta ^{2}+\frac{1}{\zeta ^{2}}=(\zeta +\frac{1}{\zeta})^{2}-2=(2x)^{2}-2=4x^{2}-2$,
$\zeta ^{3}+\frac{1}{\zeta ^{3}}=(\zeta +\frac{1}{\zeta})^{3}-3(\zeta +\frac{1}{\zeta})=(2x)^{3}-3(2x)=8x^{3}-6x$
and we get the equation $8x^{3}-4x^{2}-4x+1=0$.
$\blacksquare\;\blacksquare$
."
Amer. Math. Monthly 100, 471-474, 1993. 
