已知 $y_1 = \sin x - 2$ 且 $y_2 = -3 \cos 2x$。

1. Determine the amplitude and period of $y_1 = \sin x - 2$

For $y_1 = 1\sin(1x) - 2$:
Amplitude $= |1| = 1$
Period $= \frac{360^\circ}{1} = 360^\circ$

2. Determine the amplitude and period of $y_2 = -3 \cos 2x$

For $y_2 = -3\cos(2x)$:
Amplitude $= |-3| = 3$
Period $= \frac{360^\circ}{2} = 180^\circ$

1. Set up the equation and apply the double angle identity for cosine

$\sin x - 2 = -3 \cos 2x$
$\sin x - 2 = -3(1 - 2\sin^2 x)$

2. Rearrange into a quadratic equation in terms of $\sin x$

$\sin x - 2 = -3 + 6\sin^2 x$
$6\sin^2 x - \sin x - 1 = 0$

3. Factorise and solve for $\sin x$

$(3\sin x + 1)(2\sin x - 1) = 0$
$\sin x = -\frac{1}{3} \text{ or } \sin x = \frac{1}{2}$

4. Find the values of $x$ in the given interval

For $\sin x = \frac{1}{2}$:
Basic angle $\alpha = 30^\circ$
$x = 30^\circ, 180^\circ - 30^\circ = 150^\circ$

For $\sin x = -\frac{1}{3}$:
Basic angle $\alpha = \sin^{-1}(\frac{1}{3}) \approx 19.471^\circ$
$x = 180^\circ + 19.471^\circ = 199.471^\circ \approx 199.5^\circ$
$x = 360^\circ - 19.471^\circ = 340.529^\circ \approx 340.5^\circ$

1. Identify key features of $y_1 = \sin x - 2$

Amplitude = 1, Period = $360^\circ$. The principal axis is $y = -2$. Maximum value = $-1$, minimum value = $-3$. Key points at $x = 0^\circ, 90^\circ, 180^\circ, 270^\circ, 360^\circ$ are $(0, -2), (90, -1), (180, -2), (270, -3), (360, -2)$.

2. Identify key features of $y_2 = -3 \cos 2x$

Amplitude = 3, Period = $180^\circ$. The principal axis is $y = 0$. Maximum value = $3$, minimum value = $-3$. It is an inverted cosine graph. Key points for one cycle at $x = 0^\circ, 45^\circ, 90^\circ, 135^\circ, 180^\circ$ are $(0, -3), (45, 0), (90, 3), (135, 0), (180, -3)$. The second cycle repeats these values up to $360^\circ$.

3. Sketch both graphs on the same axes

Draw the axes for $0^\circ \le x \le 360^\circ$ and $-3 \le y \le 3$. Plot the key points for $y_1$ and draw a smooth sine curve. Plot the key points for $y_2$ and draw two cycles of a smooth inverted cosine curve. Ensure the intersections align with the solutions from part (ii).

1. Identify the intersection points from part (ii)

From part (ii), the solutions to $y_1 = y_2$ are $x = 30^\circ, 150^\circ, 199.5^\circ, 340.5^\circ$.

2. Use the graphs from part (iii) to determine where $y_2 > y_1$

The inequality $y_2 - y_1 > 0$ is equivalent to $y_2 > y_1$.
From the sketch in part (iii), the curve $y_2 = -3 \cos 2x$ lies above the curve $y_1 = \sin x - 2$ between the first and second intersection points, and between the third and fourth intersection points.
Therefore, the required set of values is $30^\circ < x < 150^\circ$ or $199.5^\circ < x < 340.5^\circ$.