= Solution
Put $u=\sin\theta$ and move all terms to one side. The <polynomial> factors as
$$
u^3-u^2-\frac u4+\frac14=(u-1)\left(u-\frac12\right)\left(u+\frac12\right).
$$
The root $u=1$ yields the already supplied $\theta=\pi/2$. In the stated <interval>, $u=1/2$ gives $\pi/6,5\pi/6$, while $u=-1/2$ gives $-\pi/6,-5\pi/6$. Thus all the other solutions are
$$
\boxed{\theta=-\frac{5\pi}{6},\ -\frac\pi6,\ \frac\pi6,\ \frac{5\pi}{6}.}
$$
There are no endpoint solutions because $\sin(\pm\pi)=0$ is not one of the three roots.
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