5. Given that $\alpha$ is an acute angle. Then the situation of the roots of the equation about $x$
$$
x^{3}-x^{2}+(\sin \alpha-3) x+1=0
$$

is ( ).
(A) Only one positive root
(B) Three positive roots
(C) One positive root, two negative roots
(D) Two positive roots, one negative root
5. D.

Obviously, $x \neq 0$. Thus, the original equation can be rewritten as $x^{2}-x+\sin \alpha-3=-\frac{1}{x}$. Let $y_{1}=x^{2}-x+\sin \alpha-3, y_{2}=-\frac{1}{x}$.

The graphs are shown in Figure 4.

The vertex of the parabola $y_{1}$ is
$$
A\left(\frac{1}{2}, \sin \alpha-\frac{13}{4}\right) .
$$

When $x=\frac{1}{2}$, $y_{2}=-2>\sin \alpha-\frac{13}{4}$, which means point $B\left(\frac{1}{2},-2\right)$ lies on the hyperbola $y_{2}=-\frac{1}{x}$, and point $B$ is directly above point $A$.

Therefore, the parabola and the hyperbola intersect at three different points, two of which have positive x-coordinates, and one has a negative x-coordinate. These three x-coordinates are all solutions to the original equation.