Adriaan van Roomen proposed a 45th degree equation in 1593(see this book, picture reference as follows):
$$
\begin{gathered}
f(x) = x^{45} - 45x^{43} + 945x^{41} - 12300x^{39} + 111150x^{37} - \color{red}{740459}x^{35} + 3764565x^{33} \\- 14945040x^{31} + 46955700x^{29} - 117679100x^{27} + 236030652x^{25} - 378658800x^{23} \\+ 483841800x^{21} - 488494125x^{19} + 384942375x^{17} - 232676280x^{15} + 105306075x^{13}
\\ - \color{red}{34512074}x^{11} + 7811375x^9 - 1138500x^7 + 95634x^5 - 3795x^3 +45x \\ = \sqrt{\frac{7}{4}-\frac{\sqrt{5}}{4}- \sqrt{\frac{15-3\sqrt{5}}{8}}} \approx 0.4158234.
\end{gathered}\tag{1}
$$
Another source said the right side is: $\sqrt{\frac{7}{4}-\frac{\sqrt{5}}{4}-\frac{5\sqrt{3}}{8}} $. ( EDIT: According to the answer, the coefficients in red are wrong.)
Another mathematician Viète noticed that left side is nothing but the expansion of sine using $$ \sin3\theta = 3\sin\theta - 4\sin^3\theta, $$ recursively, and the equation (1) reduced to: $$ \sin(\alpha) = c, \quad \text{ for } \alpha \in (0,\frac{\pi}{2}) \tag{2} $$ where $x = \sin(\alpha/45)$ (EDIT: According to the answer, this is not correct, should be $x = 2\sin(\alpha/45)$).
Now $2k\pi+\alpha$ satisfies (2) too, hence the solutions are: $$ x = \sin\left(\frac{\alpha + 2k\pi}{45}\right), \quad \text{ for } k = 0,1,2,\ldots,44. $$ Totally 45 roots in $(-1,1)$.
However, when drawing the graph of $f(x)-c$, something does not look quite right:


There are two other roots near roughly 3 and -3? Am I missing something here?
Typos! Scroll to the end to see which coefficients are typos.
$$\begin{align} \sin(45t)&=\Im\left(e^{45it}\right)\\ &=\Im\left(\left(e^{it}\right)^{45}\right)\\ &=\Im\left(\left(\cos(t)+i\sin(t)\right)^{45}\right)\\ &=\Im\left(\sum_{k=0}^{45}\binom{45}{k}\left(i\sin(t)\right)^k\cos^{45-k}(t)\right)\\ &=\frac{1}{i}\left(\sum_{k=0}^{22}\binom{45}{2k+1}\left(i\sin(t)\right)^{2k+1}\cos^{44-2k}(t)\right)\\ &=\sum_{k=0}^{22}\binom{45}{2k+1}(-1)^k\sin^{2k+1}(t)\cos^{44-2k}(t)\\ &=\sum_{k=0}^{22}\binom{45}{2k+1}(-1)^k\sin^{2k+1}(t)\left(1-\sin^2(t)\right)^{22-k}\\ &=\sum_{k=0}^{22}\binom{45}{2k+1}(-1)^k\sin^{2k+1}(t)\sum_{j=0}^{22-k}\binom{22-k}{j}(-1)^j\sin^{2j}(t)\\ &=\sum_{k=0}^{22}\sum_{j=0}^{22-k}\binom{45}{2k+1}\binom{22-k}{j}(-1)^{j+k}\sin^{2k+2j+1}(t)\\ &=\sum_{k=0}^{22}\sum_{j=k}^{22}\binom{45}{2k+1}\binom{22-k}{j-k}(-1)^{j}\sin^{2j+1}(t)\\ &=\sum_{j=0}^{22}\sum_{k=0}^{j}\binom{45}{2k+1}\binom{22-k}{j-k}(-1)^{j}\sin^{2j+1}(t)\\ &=\sum_{j=0}^{22}(-1)^{j}\left[\sum_{k=0}^{j}\binom{45}{2k+1}\binom{22-k}{j-k}\right]\sin^{2j+1}(t)\\ &=\sum_{j=0}^{22}(-1)^{j}\left[\sum_{k=0}^{j}\binom{45}{2k+1}\binom{22-k}{j-k}\right]x^{2j+1}\\ &=45x+\cdots+17592186044415x^{45} \end{align}$$
But now if we replace $x$ by $\frac{x}{2}$ and multiply the whole thing by $2$, we have
$$ \sum_{j=0}^{22}\left(\frac{-1}{4}\right)^{j}\left[\sum_{k=0}^{j}\binom{45}{2k+1}\binom{22-k}{j-k}\right]x^{2j+1} $$ which expands as $$ \begin{gathered} f(x) = x^{45} - 45x^{43} + 945x^{41} - 12300x^{39} + 111150x^{37} - 740\mathbf{2}59x^{35} + 3764565x^{33} \\- 14945040x^{31} + 46955700x^{29} - 117679100x^{27} + 236030652x^{25} - 378658800x^{23} \\+ 483841800x^{21} - 488494125x^{19} + 384942375x^{17} - 232676280x^{15} + 105306075x^{13} \\ - 3451207\mathbf{5}x^{11} + 7811375x^9 - 1138500x^7 + 95634x^5 - 3795x^3 +45x \\ \end{gathered} $$ and the coefficients of $x^{35}$ and of $x^{11}$ are typos.
Note that $x=2\sin(\alpha/45)$ is the substitution, not $x=\sin(\alpha/45)$, and that we multiplied by $2$, so that doubles the value of $c$.