Solution
= Solution
Take the imaginary part of <De Moivre's theorem>, $(\cos\theta+i\sin\theta)^5=\cos5\theta+i\sin5\theta$. With $s=\sin\theta$ and $\cos^2\theta=1-s^2$,
$$
\sin5\theta=5(1-s^2)^2s-10(1-s^2)s^3+s^5=\boxed{16s^5-20s^3+5s}.
$$
The required equation cancels the fifth powers and becomes $5s(1-4s^2)=0$. Thus $\sin\theta=0,\pm1/2$, giving all real solutions
$$
\boxed{\theta=n\pi\quad\text{or}\quad\theta=n\pi\pm\frac\pi6,\qquad n\in\mathbb Z.}
$$