写出函数 $y=3\sin\frac{x}{2}+1$ 图像的振幅和周期。 画出 $y=3\sin\frac{x}{2}+1$ 在 $0 \le x \le 4\pi$ 范围内的图像草图。 通过在草图上添加另一条合适的曲线，确定方程 $\sin\frac{x}{2} = -\frac{1}{3} - \frac{2}{3}\cos\frac{x}{2}$ 的解的个数。

1. Determine the amplitude of the function

The amplitude is given by the coefficient of the sine term.
Amplitude = $|3| = 3$

2. Determine the period of the function

The period of $y = a\sin(bx) + c$ is $\frac{2\pi}{b}$.
Here, $b = \frac{1}{2}$.
Period = $\frac{2\pi}{1/2} = 4\pi$

1. Determine the key coordinates of the graph for one period

From part (a), the period is $4\pi$ and the amplitude is $3$. The principal axis is $y = 1$.

• Start point: $(0, 1)$
• Maximum point: $(\pi, 1+3) = (\pi, 4)$
• Midpoint (crossing principal axis): $(2\pi, 1)$
• Minimum point: $(3\pi, 1-3) = (3\pi, -2)$
• End point: $(4\pi, 1)$

2. Sketch the curve passing through the key points

Draw a smooth sine curve passing through $(0, 1)$, $(\pi, 4)$, $(2\pi, 1)$, $(3\pi, -2)$, and $(4\pi, 1)$.

1. Manipulate the given equation to form the expression of the original graph on one side.

$\sin\frac{x}{2} = -\frac{1}{3} - \frac{2}{3}\cos\frac{x}{2}$
Multiply by 3:
$3\sin\frac{x}{2} = -1 - 2\cos\frac{x}{2}$
Add 1 to both sides to match the original graph $y = 3\sin\frac{x}{2} + 1$:
$3\sin\frac{x}{2} + 1 = -2\cos\frac{x}{2}$

2. Identify the suitable curve to draw.

The suitable curve to draw is $y = -2\cos\frac{x}{2}$.

3. Determine the number of intersections between the two graphs in the given domain.

The graph of $y = 3\sin\frac{x}{2} + 1$ completes one full cycle in $0 \le x \le 4\pi$, starting at $(0, 1)$, peaking at $(\pi, 4)$, crossing the midline at $(2\pi, 1)$, reaching a minimum at $(3\pi, -2)$, and ending at $(4\pi, 1)$.
The graph of $y = -2\cos\frac{x}{2}$ also completes one full cycle in $0 \le x \le 4\pi$, starting at $(0, -2)$, crossing the x-axis at $(\pi, 0)$, peaking at $(2\pi, 2)$, crossing the x-axis at $(3\pi, 0)$, and ending at $(4\pi, -2)$.
Sketching these two curves on the same axes shows they intersect at exactly 2 points.
Therefore, there are 2 solutions.