Sketch the graphs of $y=-2\cos 3x+1$ and $y= -\dfrac{2}{\pi}x+1$ on the same axes, for $0 \le x \le \pi$. Explain how the number of solutions for $2\cos 3x = \dfrac{2}{\pi}x$ can be found using the graphs.

1. Determine the key features of the trigonometric graph $y = -2\cos 3x + 1$

Amplitude $= |-2| = 2$
Period $= \frac{2\pi}{3}$
Principal axis: $y = 1$
Maximum value $= 1 + 2 = 3$
Minimum value $= 1 - 2 = -1$

2. Calculate key coordinates for $y = -2\cos 3x + 1$ in the interval $0 \le x \le \pi$

At $x = 0$, $y = -2(1) + 1 = -1$
At $x = \frac{\pi}{6}$, $y = -2(0) + 1 = 1$
At $x = \frac{\pi}{3}$, $y = -2(-1) + 1 = 3$
At $x = \frac{\pi}{2}$, $y = -2(0) + 1 = 1$
At $x = \frac{2\pi}{3}$, $y = -2(1) + 1 = -1$
At $x = \frac{5\pi}{6}$, $y = -2(0) + 1 = 1$
At $x = \pi$, $y = -2(-1) + 1 = 3$

3. Determine the key features of the linear graph $y = -\frac{2}{\pi}x + 1$

This is a straight line.
At $x = 0$, $y = 1$
At $x = \pi$, $y = -\frac{2}{\pi}(\pi) + 1 = -2 + 1 = -1$

4. Sketch both graphs on the same axes

Draw the axes with $x$ from $0$ to $\pi$ and $y$ from $-1$ to $3$.
Plot the points for the cosine curve and draw a smooth curve through them (1.5 cycles).
Plot the points $(0, 1)$ and $(\pi, -1)$ and draw a straight line connecting them.

1. Rearrange the given equation to match the equations of the graphs drawn in part (a)

Given equation: $2\cos 3x = \frac{2}{\pi}x$
Multiply by $-1$:
$-2\cos 3x = -\frac{2}{\pi}x$
Add $1$ to both sides:
$-2\cos 3x + 1 = -\frac{2}{\pi}x + 1$

2. State the relationship between the rearranged equation and the graphs

The left side is the equation of the curve $y = -2\cos 3x + 1$ and the right side is the equation of the line $y = -\frac{2}{\pi}x + 1$. The number of solutions is the number of points of intersection of these two graphs.