Four, (20 points) Solve the equation for $x$
$$
\left(\cos ^{2} \frac{\theta}{2}\right) x^{3}+\left(3 \cos ^{2} \frac{\theta}{2}-4\right) x+\sin \theta=0 .
$$
If $\cos \frac{\theta}{2}=0$, then $\sin \theta=0$. Hence $x=0$.
If $\sin \frac{\theta}{2}=0$, then $\cos ^{2} \frac{\theta}{2}=1, \sin \theta=0$.
Thus, $x=0$ or $\pm 1$.
If $\cos \frac{\theta}{2} \neq 0$, and $\sin \frac{\theta}{2} \neq 0$, then $x \neq 0$, and the original equation becomes
$$
\begin{array}{l}
x^{3}-\left(3-4 \sec ^{2} \frac{\theta}{2}\right) x+2 \tan \frac{\theta}{2}=0 \\
\Rightarrow x^{3}-\left(4 \tan ^{2} \frac{\theta}{2}+1\right) x+2 \tan \frac{\theta}{2}=0 \\
\Rightarrow 4 x \tan ^{2} \frac{\theta}{2}-2 \tan \frac{\theta}{2}+x-x^{3}=0 \\
\Rightarrow \tan \frac{\theta}{2}=\frac{x}{2} \text { or } \frac{1-x^{2}}{2 x} \\
\Rightarrow x=2 \tan \frac{\theta}{2} \text { or }-\tan \frac{\theta}{2} \pm \sec \frac{\theta}{2} .
\end{array}
$$

In summary, the solutions to the original equation are:
When $\cos \frac{\theta}{2}=0$, i.e., $\theta=(2 k+1) \pi(k \in \mathbf{Z})$, $x=0$;

When $\cos \frac{\theta}{2} \neq 0$, i.e., $\theta \neq(2 k+1) \pi(k \in \mathbf{Z})$, $x=2 \tan \frac{\theta}{2}$ or $-\tan \frac{\theta}{2} \pm \sec \frac{\theta}{2}$ (the case where $\sin \frac{\theta}{2}=0$ is included in this).