In the diagram, the graphs of $f(x) = ext{cos}(x + a)$ and $g(x) = ext{sin} 2x$ are drawn for the interval $x \in [-180^{\circ}; 180^{\circ}]$ - NSC Mathematics - Question 6 - 2024 - Paper 2

The amplitude of the function $g(x) = \text{sin} 2x$ is 1, as the sine function fluctuates between -1 and 1.

Given that $a = -45^{\circ}$, we can state that the value of $a$ is $-45^{\circ}$.

To find $k$, we can analyze the intersection points:

For point $N(-75^{\circ}; k)$: $k = \text{sin}(2 \cdot (-75^{\circ})) = \text{sin}(-150^{\circ}) = -\text{sin}(30^{\circ}) = -\frac{1}{2}$

For point $Q(165^{\circ}; k)$: $k = \text{sin}(2 \cdot 165^{\circ}) = \text{sin}(330^{\circ}) = -\text{sin}(30^{\circ}) = -\frac{1}{2}$

So, $k = -\frac{1}{2}$ for both points.

$g(x + 60^{\circ}) = \text{sin}(2(x + 60^{\circ}))$ and $f(x + 60^{\circ}) = \text{cos}(x + 60^{\circ} - 45^{\circ}) = \text{cos}(x + 15^{\circ})$.

Set the equations equal: $\text{sin}(2x + 120^{\circ}) = \text{cos}(x + 15^{\circ})$

After manipulating and finding common solutions, we discover: $x = -15^{\circ}$

Rearranging gives: $\sqrt{2} \text{sin} 2x - \text{sin} x - \cos x = 0$

Using special angles and behavior of the sine and cosine functions, we analyze the intervals within $[-90^{\circ}; 90^{\circ}]$. After evaluating equations, we determine there are 2 roots in this interval.