(11 (20 points) Solve the equation for $x$: $\left(\cos ^{2} \frac{\theta}{2}\right) x^{3}+\left(3 \cos ^{2} \frac{\theta}{2}-4\right) x+$ $\sin \theta=0$.
11 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 $x= \pm 1$.
If $\cos \frac{\theta}{2} \neq 0$, and $\sin \frac{\theta}{2} \neq 0$, then the original equation becomes
$$
x^{3}+\left(3-4 \sec ^{2} \frac{\theta}{2}\right) x+2 \tan \frac{\theta}{2}=0,
$$

which is
$$
x^{3}-\left(4 \tan ^{2} \frac{\theta}{2}+1\right) x+2 \tan \frac{\theta}{2}=0,
$$

which is
$$
4 x \tan ^{2} \frac{\theta}{2}-2 \tan \frac{\theta}{2}+x-x^{3}=0 .
$$

Since $\sin \frac{\theta}{2} \neq 0$, then $\tan \frac{\theta}{2} \neq 0$, thus $x \neq 0$.
Therefore, $4 x \tan ^{2} \frac{\theta}{2}-2 \tan \frac{\theta}{2}+x-x^{3}=0$ is a quadratic equation in $\tan \frac{\theta}{2}$, solving it yields $\tan \frac{\theta}{2}=\frac{x}{2}$, or $\tan \frac{\theta}{2}=\frac{1-x^{2}}{2 x}$.

Solving these two equations for $x$ gives $x=2 \tan \frac{\theta}{2}$, or $x=-\tan \frac{\theta}{2} \pm \sec \frac{\theta}{2}$.
In summary, the solutions to the original equation are:
When $\cos \frac{\theta}{2}=0$, i.e., $\theta=2 k \pi+\pi(k \in \mathbf{Z})$, $x=0$;
When $\cos \frac{\theta}{2} \neq 0$, i.e., $\theta \neq 2 k \pi+\pi(k \in \mathbf{Z})$, $x=2 \tan \frac{\theta}{2}$, or $x=$ $-\tan \frac{\theta}{2} \pm \sec \frac{\theta}{2}$ (the case when $\sin \frac{\theta}{2}=0$ is included in this).