4. 2. 35 * Given Given that the equation $x^{3} \sin \theta-(\sin \theta+2) x^{2}+6 x-4=0$ 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.
The original equation is $(x-1)\left(x^{2} \sin \theta-2 x+4\right)=0$, so $x_{1}=1$, and the quadratic equation in $x$, $x^{2} \sin \theta-2 x+4=0$, has two positive real roots. Therefore, $\Delta=4-16 \sin \theta \geqslant 0$, which implies $\sin \theta \leqslant \frac{1}{4}$. By Vieta's formulas, we know that $\sin \theta>0$, hence $0<\sin \theta \leqslant \frac{1}{4}$.
Since $9 \sin ^{2} \theta-4 \sin \theta+3=9\left(\sin \theta-\frac{2}{9}\right)^{2}+\frac{23}{9} \geqslant \frac{23}{9}$, and
$$
\begin{aligned}
& (1-\cos \theta)(2 \cos \theta-6 \sin \theta-3 \sin 2 \theta+2) \\
= & 2(1-\cos \theta)(1+\cos \theta)(1-3 \sin \theta)=2 \sin ^{2} \theta(1-3 \sin \theta) \\
= & \frac{8}{9} \cdot \frac{3}{2} \sin \theta \cdot \frac{3}{2} \sin \theta \cdot(1-3 \sin \theta) \\
\leqslant & \frac{8}{9} \cdot\left(\frac{\frac{3}{2} \sin \theta+\frac{3}{2} \sin \theta+1-3 \sin \theta}{3}\right)^{3}=\frac{8}{9 \times 27},
\end{aligned}
$$

Therefore, $u \geqslant \frac{\frac{23}{9}}{\frac{8}{9 \times 27}}=\frac{23 \times 27}{8}=\frac{621}{8}$, with equality when $\sin \theta=\frac{2}{9}$. Thus, the minimum value of $u$ is
$$
\frac{621}{8} .
$$