= Solution
By <De Moivre's theorem>, $\sin5\theta$ is the <imaginary part> of $(\cos\theta+i\sin\theta)^5$. The <binomial theorem> gives, with $s=\sin\theta$,
$$
\sin5\theta=5\cos^4\theta\,s-10\cos^2\theta\,s^3+s^5.
$$
Use $\cos^2\theta=1-s^2$ and simplify:
$$
\boxed{\sin5\theta=16s^5-20s^3+5s.}
$$
The requested equation cancels the $16s^5$ term, leaving $5s(1-4s^2)=0$. Thus $s=0$ or $s=\pm1/2$. The complete list in $0\leq\theta<2\pi$ is
$$
\boxed{\theta=0,\ \frac\pi6,\ \frac{5\pi}6,\ \pi,\ \frac{7\pi}6,\ \frac{11\pi}6.}
$$
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