1恻9. 筈不等式
$$
3 \sqrt{1-\frac{x^{2}}{4}}<\frac{1}{2} x+2
$$

1 side 9. Inequality
$$
3 \sqrt{1-\frac{x^{2}}{4}}<\frac{1}{2} x+2
$$
We use the mixed system
$$
y<\frac{1}{2} x+2, y^{2}=9\left(1-\frac{x^{2}}{4}\right).
$$
$y \geqslant 0$ to replace the original inequality. The equation
$y^{2}=9\left(1-\frac{x^{2}}{4}\right)$ represents an ellipse (Figure 4).

The solution sought is the part of the ellipse that lies below the line $y=\frac{1}{2} x+2$,
within the half-plane $y \geqslant 0$. The x-coordinates of the points on these two segments (bold parts) are the solutions to the original inequality.
The x-coordinates of the intersection points of the ellipse and the line are determined by the system of equations
$$
\left\{\begin{array}{l}
y^{2}=9\left(1-\frac{x^{2}}{4}\right), \\
y=\frac{1}{2} x+2
\end{array}\right.
$$

Eliminating $y$, we get
$$
\begin{array}{l}
5 x^{2}+4 x-10=0, \\
x_{1}=\frac{-2+3 \sqrt{6}}{5} \text{ (x-coordinate of point A), } \\
x_{2}=\frac{-2-3 \sqrt{6}}{5} \text{ (x-coordinate of point B) } \\
\text{Thus }-2 \leqslant x<\frac{-2-3 \sqrt{6}}{5}, \frac{-2+3 \sqrt{6}}{5}
\end{array}
$$
$<x \leqslant 2$ is the solution to the original inequality.