The function $f$ is defined, for $0 \le x \le 2\pi$, by $f(x) = 1 - 2 \cos 2x$.

1. Determine the period of the function

The function is $f(x) = 1 - 2 \cos 2x$. The coefficient of $x$ is $2$.
$\text{Period} = \frac{2\pi}{2} = \pi$

2. Determine the amplitude of the function

The coefficient of the cosine term is $-2$.
$\text{Amplitude} = |-2| = 2$

1. Determine the maximum and minimum values of the graph

The maximum value of $-2\cos 2x$ is $2$, so the maximum value of $f(x)$ is $1 + 2 = 3$.
The minimum value of $-2\cos 2x$ is $-2$, so the minimum value of $f(x)$ is $1 - 2 = -1$.

2. Identify the number of cycles and key $x$-coordinates

The period is $\pi$, so there are two full cycles in the interval $0 \le x \le 2\pi$.
The key points for the first cycle occur at intervals of $\frac{\pi}{4}$: $0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi$.

3. Sketch the graph

Plot the key points:
$(0, -1), (\frac{\pi}{4}, 1), (\frac{\pi}{2}, 3), (\frac{3\pi}{4}, 1), (\pi, -1)$
$(\frac{5\pi}{4}, 1), (\frac{3\pi}{2}, 3), (\frac{7\pi}{4}, 1), (2\pi, -1)$
Draw a smooth curve through these points.

1. Identify the type of graph and find key points to plot

The equation $y = \frac{4x}{\pi} - 2$ represents a straight line.
When $x = 0$, $y = -2$.
When $x = \frac{\pi}{2}$, $y = \frac{4(\pi/2)}{\pi} - 2 = 0$.
When $x = \pi$, $y = \frac{4\pi}{\pi} - 2 = 2$.
When $x = 2\pi$, $y = \frac{4(2\pi)}{\pi} - 2 = 6$.

2. Sketch the line on the same diagram

Draw a straight line passing through the points $(0, -2)$, $(\frac{\pi}{2}, 0)$, $(\pi, 2)$, and $(2\pi, 6)$ on the axes from part (b).

1. Rearrange the given equation to match the functions sketched in parts (b) and (c).

$2 \cos 2x - 3 + \frac{4x}{\pi} = 0$
$-2 \cos 2x = \frac{4x}{\pi} - 3$
$1 - 2 \cos 2x = \frac{4x}{\pi} - 2$

2. Relate the rearranged equation to the number of intersections of the graphs.

The equation is now in the form $f(x) = \frac{4x}{\pi} - 2$. The number of solutions is the number of points of intersection between the curve $y = 1 - 2 \cos 2x$ and the line $y = \frac{4x}{\pi} - 2$ for $0 \le x \le 2\pi$.

3. Determine the number of intersections from the properties of the graphs.

Checking key points:
At $x = 0$, curve $y = -1$, line $y = -2$ (curve is above line).
At $x = \frac{\pi}{2}$, curve $y = 3$, line $y = 0$ (curve is above line).
At $x = \frac{3\pi}{4}$, curve $y = 1 - 2\cos\left(\frac{3\pi}{2}\right) = 1$, line $y = \frac{4}{\pi}\left(\frac{3\pi}{4}\right) - 2 = 1$. They intersect exactly here.
At $x = \pi$, curve $y = -1$, line $y = 2$ (line is above curve).
For $x > \frac{3\pi}{4}$, the line continues to rise (up to $y=6$ at $x=2\pi$) while the curve's maximum is $3$, so they do not intersect again.
Thus, there is exactly $1$ point of intersection.