Solve $\sin(5\theta)=1$, $0<\theta<2\pi$. Show that the roots of $16x^4+16x^3-4x^2-4x+1=0$ are $x=\sin{\frac{(4r+1)\pi}{10}}$, $r=0,2,3,4$. This is a very interesting problem that I came across. I know it's got something to do with trigonometry identities, polynomials and complex numbers, but other than that, I'm not too sure how to approach this. Any guidance hints or help would be greatly appreciated. Thanks in advance. Here's the problem:

Solve the equation $\sin(5\theta)=1$ for $0<\theta<2\pi$ and hence show that the roots of the equation $16x^4+16x^3-4x^2-4x+1=0$ are $x=\sin{\frac{(4r+1)\pi}{10}}$, where for $r=0,2,3,4$. Determine the exact value of $\sin\frac\pi{10}\sin\frac{3\pi}{10}$.

It might be useful to find $\sin(5\theta)$ in terms of $sin^n \theta$ and so on.

Use$$\sin5\theta=\sin{x}\cos4\theta+\cos{x}\sin4\theta=\sin{\theta}(2\cos^22\theta-1)+4\cos^2\theta\cos2x\sin{\theta}=$$
$$=\sin{\theta}(2\cos^22\theta-1+2(1+\cos2\theta)\cos2\theta))=\sin{\theta}(4\cos^22\theta+2\cos2\theta-1)$$
and $16x^4+16x^3-4x^2-4x+1=(4x^2+2x-1)^2$
