It is given that $\text{f}(x) = 2 - 3\sin 3x$ and $\text{g}(x) = 4\cos\left(\frac{x}{2}\right) - 1$.

1. State the range of the sine function

We know that $-1 \le \sin 3x \le 1$.

2. Determine the least and greatest values of f(x)

Multiply by $-3$:
$3 \ge -3\sin 3x \ge -3$
Add 2:
$5 \ge 2 - 3\sin 3x \ge -1$
So, the least value is $-1$ and the greatest value is $5$.

1. Determine the range of the cosine component

The value of $\cos\left(\frac{x}{2}\right)$ ranges from $-1$ to $1$. Therefore, $4\cos\left(\frac{x}{2}\right)$ ranges from $-4$ to $4$.

2. Calculate the least and greatest values of $\text{g}(x)$

Greatest value of $\text{g}(x) = 4(1) - 1 = 3$
Least value of $\text{g}(x) = 4(-1) - 1 = -5$

1. Calculate the period of the function

$\text{Period} = \frac{360^\circ}{3} = 120^\circ$

1. Calculate the period of the function

The period of $\cos(bx)$ is given by $\frac{360^\circ}{b}$.
For $\text{g}(x) = 4\cos\left(\frac{x}{2}\right) - 1$, $b = \frac{1}{2}$.
$\text{Period} = \frac{360^\circ}{\frac{1}{2}} = 720^\circ$

1. Determine the key features of $y = \text{f}(x) = 2 - 3\sin 3x$ for $0^\circ \le x \le 180^\circ$

Amplitude = 3, Principal axis: $y = 2$.
Period = $\frac{360^\circ}{3} = 120^\circ$.
Max value = $2 - 3(-1) = 5$, Min value = $2 - 3(1) = -1$.
Starts at $(0^\circ, 2)$, goes down to $(30^\circ, -1)$, crosses axis at $(60^\circ, 2)$, reaches max at $(90^\circ, 5)$, crosses axis at $(120^\circ, 2)$, goes down to $(150^\circ, -1)$, and ends at $(180^\circ, 2)$.

2. Determine the key features of $y = \text{g}(x) = 4\cos\left(\frac{x}{2}\right) - 1$ for $0^\circ \le x \le 180^\circ$

Amplitude = 4, Principal axis: $y = -1$.
Period = $\frac{360^\circ}{1/2} = 720^\circ$.
Max value = $4(1) - 1 = 3$, Min value = $4(-1) - 1 = -5$.
Starts at $(0^\circ, 3)$, decreases to $(180^\circ, -1)$ since $180^\circ$ is a quarter of the period.

3. Sketch both graphs on the same axes

Draw axes with $x$ from $0^\circ$ to $180^\circ$ and $y$ from $-1$ to $5$.
Plot the key points for $y = \text{f}(x)$ and draw a smooth inverted sine curve completing 1.5 cycles.
Plot the key points for $y = \text{g}(x)$ and draw a smooth cosine curve completing 0.25 cycles.
Label both graphs clearly.

1. Rearrange the given equation to relate it to the functions $\text{f}(x)$ and $\text{g}(x)$.

$2\cos\frac{x}{2} + \frac{3}{2}\sin 3x = \frac{3}{2}$
Multiply the entire equation by 2:
$4\cos\frac{x}{2} + 3\sin 3x = 3$
Rearrange the terms to form $\text{f}(x)$ and $\text{g}(x)$ on either side:
$4\cos\frac{x}{2} - 1 = 2 - 3\sin 3x$
$\text{g}(x) = \text{f}(x)$

2. Determine the number of solutions by finding the number of intersection points of the two graphs.

The number of solutions to the equation is equal to the number of points of intersection between the graphs of $y = \text{f}(x)$ and $y = \text{g}(x)$ for $0^\circ \le x \le 180^\circ$.
From the sketch in part (v), the curve $y = \text{g}(x)$ intersects the curve $y = \text{f}(x)$ at exactly 3 points in the given interval.