In the diagram, the graphs of $f(x) = -3 \sin \left( \frac{x}{2} \right)$ and $g(x) = 2\cos \left( x - 60^{\circ} \right)$ are drawn in the interval $x \in [-180^{\circ}; 180^{\circ}]$ - NSC Mathematics - Question 6 - 2018 - Paper 2

The function $g(x) = 2\cos \left( x - 60^{\circ} \right)$ has a maximum value of $2$ and a minimum value of $-2$. Therefore, the range of $g$ is:

$\text{Range of } g = [-2, 2].$

To calculate $f(p) - g(p)$, we first need to evaluate both functions at $p$. Assume $p$ is a specific angle in degrees. We substitute $p$ into both functions:

1. Calculate: $f(p) = -3 \sin \left( \frac{p}{2} \right)$
2. Calculate: $g(p) = 2\cos \left( p - 60^{\circ} \right)$

Thus, $f(p) - g(p) = -3 \sin \left( \frac{p}{2} \right) - 2\cos \left( p - 60^{\circ} \right)$.

From the graph of $g(x)$, we observe that $g(x) > 0$ in the intervals where the cosine function is positive. The cosine function is positive in the following intervals within the specified range:

• From $-60^{\circ}$ to $60^{\circ}$ Thus, the values of $x$ for which $g(x) > 0$ are:

$x \in (-60^{\circ}, 60^{\circ}).$

From the graph of $f(x)$, we need to identify the portions where the sine function is positive. The sine function is positive in the intervals:

• From $0^{\circ}$ to $180^{\circ}$. Thus, the values of $x$ for which $f(x) > 0$ are:

$x \in (0^{\circ}, 180^{\circ}).$