In the diagram, the graphs of $f(x) = 2 ext{sin}2x$ and $g(x) = - ext{cos}(x + 45^{ ext{o}})$ are drawn for the interval $x ext{ e } [0^{ ext{o}}; 180^{ ext{o}}]$ - NSC Mathematics - Question 6 - 2023 - Paper 2

To find the range of $g(x) = - ext{cos}(x + 45^{ ext{o}})$, we note that the cosine function has a range of $[-1, 1]$. Thus, for $g(x)$:

$g(x) ext{ will have a range of } [-1, 1] ext{ inverting yields } [-1, 0]$

This gives us g(x) ext{ e } [0, rac{ ext{√2}}{2}].

For the condition $f(x) \cdot g(x) > 0$:

1. Determine where both functions are positive or negative.
2. Since $f(x) = 2 ext{sin}2x > 0$ when $0^{ ext{o}} < x < 45^{ ext{o}}$.
3. Find intersections of the sine function and cosine function for given intervals. Thus, the values of $x$ satisfying the inequality are:

$x \text{ e } (45^{\text{o}}, 90^{\text{o}})$

Solving $f(x) + 1 \text{ ≤ } 0$:

$f(x) \leq -1 ext{ implies } 2 ext{sin}2x \leq -1 \Rightarrow ext{sin}2x \leq -\frac{1}{2}$

This is satisfied at:

$x \text{ e } [105^{\text{o}}, 165^{\text{o}}]$

Since $p(x) = -f(x)$, we can find:

$p(k) = -f(k) ext{ when } f(k) = 1.$

From earlier calculations, the solutions yield:

$k = 15^{\text{o}}, 75^{\text{o}}$

To translate $g(x) = - ext{cos}(x + 45^{\text{o}})$, we substitute:

$h(x) = - ext{cos}(x + 90^{\text{o}})$

Since $h$ is simplified, the equation becomes:

$h(x) = ext{sin}(x)$