Answer: The roots are \(x=\sin\left(\frac{7\pi}{30}\right),\sin\left(\frac{11\pi}{30}\right),\sin\left(\frac{31\pi}{30}\right),\sin\left(\frac{35\pi}{30}\right),\sin\left(\frac{43\pi}{30}\right)\).

Let \(z=\cos\theta+i\sin\theta\). Then by de Moivre's theorem,

\(z^5=(\cos\theta+i\sin\theta)^5=\cos 5\theta+i\sin 5\theta\).

Expanding the left-hand side and taking imaginary parts gives

\(\sin 5\theta=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta\).

Now write \(\cos^2\theta=1-\sin^2\theta\). Then

\(\sin 5\theta=5\sin\theta(1-\sin^2\theta)^2-10\sin^3\theta(1-\sin^2\theta)+\sin^5\theta\).

Expanding and simplifying:

\(\sin 5\theta=5\sin\theta(1-2\sin^2\theta+\sin^4\theta)-10\sin^3\theta+10\sin^5\theta+\sin^5\theta\)

\(=5\sin\theta-10\sin^3\theta+5\sin^5\theta-10\sin^3\theta+10\sin^5\theta+\sin^5\theta\)

\(=16\sin^5\theta-20\sin^3\theta+5\sin\theta\).

So \(\sin 5\theta=16\sin^5\theta-20\sin^3\theta+5\sin\theta\), as required.

Now let \(x=\sin\theta\). The equation

\(32x^5-40x^3+10x+1=0\)

can be written as

\(2(16x^5-20x^3+5x)+1=0\),

so using the identity just proved,

\(2\sin 5\theta+1=0\).

Hence \(\sin 5\theta=-\frac12\).

Now solve \(\sin u=-\frac12\). In \([0,2\pi)\), the solutions are

\(u=\frac{7\pi}{6},\frac{11\pi}{6}\).

So for \(u=5\theta\),

\(5\theta=\frac{7\pi}{6},\frac{11\pi}{6},\frac{19\pi}{6},\frac{23\pi}{6},\frac{31\pi}{6}\)

giving

\(\theta=\frac{7\pi}{30},\frac{11\pi}{30},\frac{19\pi}{30},\frac{23\pi}{30},\frac{31\pi}{30}\).

Therefore the roots, in the form \(\sin(q\pi)\), are

\(\sin\left(\frac{7\pi}{30}\right),\sin\left(\frac{11\pi}{30}\right),\sin\left(\frac{19\pi}{30}\right),\sin\left(\frac{23\pi}{30}\right),\sin\left(\frac{31\pi}{30}\right)\).