10. (14 points) Given the equation in terms of $x$
$$
x^{3} \sin \theta-(\sin \theta+2) x^{2}+6 x-4=0
$$

has 3 positive real roots. Find
$$
u=\frac{9 \sin ^{2} \theta-4 \sin \theta+3}{(1+\cos \theta)(2 \cos \theta-6 \sin \theta-3 \sin 2 \theta+2)}
$$

the minimum value.
10. The original equation is $(x-1)\left(x^{2} \sin \theta-2 x+4\right)=0$.

Since the original equation has 3 positive real roots, the quadratic equation in $x$, $x^{2} \sin \theta-2 x+4=0$, has 2 positive real roots, i.e., $\square$
$$
\begin{array}{l}
\left\{\begin{array}{l}
\Delta=4-16 \sin \theta \geqslant 0, \\
\sin \theta>0
\end{array} \Leftrightarrow 0<\sin \theta \leqslant \frac{1}{4} .\right. \\
\text { Also, } 9 \sin ^{2} \theta-4 \sin \theta+3 \\
=9\left(\sin \theta-\frac{2}{9}\right)^{2}+\frac{23}{9} \geqslant \frac{23}{9}, \\
0<(1-\cos \theta)(2 \cos \theta-6 \sin \theta-3 \sin 2 \theta+2) \\
=2(1-\cos \theta)(1+\cos \theta)(1-3 \sin \theta) \\
=2 \sin ^{2} \theta(1-3 \sin \theta) \\
=\frac{8}{9} \times \frac{3}{2} \sin \theta \times \frac{3}{2} \sin \theta(1-3 \sin \theta) \\
\leqslant \frac{8}{9}\left(\frac{1}{3}\right)^{3}=\frac{8}{9 \times 27} .
\end{array}
$$

Then $u \geqslant \frac{\frac{23}{9}}{\frac{8}{9 \times 27}}=\frac{621}{8}$.
When $\sin \theta=\frac{2}{9}$, the equality holds.
Thus, $u_{\min }=\frac{621}{8}$.