1. **Problem Statement:** We have two functions: $f(x) = -2 \cos x$ and $g(x) = \sin 2x$ defined on the interval $-90^\circ \leq x \leq 180^\circ$. We need to: - Draw their graphs on the same axes. - Identify intercepts and turning points. - Solve for $x$ where: a) $g(x) - f(x) = 2$ b) $f(x) \leq g(x)$ c) Both $f(x)$ and $g(x)$ are increasing. - Find the equation of $h(x)$ after reflecting $f(x)$ in the x-axis and shifting it 30° right. 2. **Graphing and Key Points:** - $f(x) = -2 \cos x$ is a cosine wave with amplitude 2, reflected about the x-axis. - $g(x) = \sin 2x$ is a sine wave with double frequency. - Domain: $-90^\circ$ to $180^\circ$. **Intercepts and turning points:** - For $f(x)$: - Intercepts where $f(x)=0$: $-2 \cos x=0 \Rightarrow \cos x=0 \Rightarrow x=90^\circ$. - Turning points where $f'(x)=0$: $f'(x) = 2 \sin x = 0 \Rightarrow x=0^\circ, 180^\circ$. - Values: $f(0) = -2$, $f(180) = 2$. - For $g(x)$: - Intercepts where $g(x)=0$: $\sin 2x=0 \Rightarrow 2x = n\pi \Rightarrow x = n \times 90^\circ$, for integers $n$. - Turning points where $g'(x)=0$: $g'(x) = 2 \cos 2x = 0 \Rightarrow 2x = 90^\circ, 270^\circ, ... \Rightarrow x=45^\circ, 135^\circ$. - Values: $g(45^\circ) = 1$, $g(135^\circ) = -1$. 3. **Solving the equations:** **a) Find $x$ where $g(x) - f(x) = 2$:** $$\sin 2x - (-2 \cos x) = 2 \Rightarrow \sin 2x + 2 \cos x = 2$$ Using double angle: $\sin 2x = 2 \sin x \cos x$, $$2 \sin x \cos x + 2 \cos x = 2$$ Factor out $2 \cos x$: $$2 \cos x (\sin x + 1) = 2$$ Divide both sides by 2: $$\cos x (\sin x + 1) = 1$$ Since $|\cos x| \leq 1$ and $|\sin x + 1| \leq 2$, equality holds when both are 1: - $\cos x = 1 \Rightarrow x=0^\circ$ - $\sin x + 1 = 1 \Rightarrow \sin x = 0$ At $x=0^\circ$, $\sin 0 = 0$, so condition holds. Check if $x=0^\circ$ satisfies original: $$\sin 0 + 2 \cos 0 = 0 + 2 = 2$$ Yes, so $x=0^\circ$ is a solution. **b) Find $x$ where $f(x) \leq g(x)$:** $$-2 \cos x \leq \sin 2x$$ Rewrite $\sin 2x$ as $2 \sin x \cos x$: $$-2 \cos x \leq 2 \sin x \cos x$$ Divide both sides by 2: $$-\cos x \leq \sin x \cos x$$ Rearranged: $$0 \leq \cos x (\sin x + 1)$$ Analyze sign of $\cos x$ and $\sin x + 1$ on $[-90^\circ, 180^\circ]$: - $\sin x + 1$ is always positive since $\sin x \geq -1$. - So inequality reduces to $\cos x \geq 0$. On $[-90^\circ, 180^\circ]$, $\cos x \geq 0$ for $x \in [-90^\circ, 90^\circ]$. Therefore, $f(x) \leq g(x)$ for $x$ in $[-90^\circ, 90^\circ]$. **c) Find $x$ where both $f(x)$ and $g(x)$ are increasing:** - $f'(x) = 2 \sin x$ (since $f(x) = -2 \cos x$) - $g'(x) = 2 \cos 2x$ For $f(x)$ increasing: $f'(x) > 0 \Rightarrow \sin x > 0$. On $[-90^\circ, 180^\circ]$, $\sin x > 0$ for $x \in (0^\circ, 180^\circ)$. For $g(x)$ increasing: $g'(x) > 0 \Rightarrow \cos 2x > 0$. $\cos 2x > 0$ when $2x \in (-90^\circ, 90^\circ)$ or $2x \in (270^\circ, 450^\circ)$. Within domain: - $2x \in (-90^\circ, 90^\circ) \Rightarrow x \in (-45^\circ, 45^\circ)$ - $2x \in (270^\circ, 450^\circ) \Rightarrow x \in (135^\circ, 225^\circ)$ but domain ends at $180^\circ$. So $g(x)$ increasing on $(-45^\circ, 45^\circ)$ and $(135^\circ, 180^\circ)$. Intersection with $f(x)$ increasing $(0^\circ, 180^\circ)$: - Overlap is $(0^\circ, 45^\circ)$ and $(135^\circ, 180^\circ)$. 4. **Equation of $h(x)$ after reflection and shift:** - Reflect $f(x)$ in x-axis: $-f(x) = 2 \cos x$. - Shift 30° right: replace $x$ by $x - 30^\circ$. So, $$h(x) = 2 \cos (x - 30^\circ)$$ **Final answers:** - a) $x = 0^\circ$ - b) $x \in [-90^\circ, 90^\circ]$ - c) $x \in (0^\circ, 45^\circ) \cup (135^\circ, 180^\circ)$ - $h(x) = 2 \cos (x - 30^\circ)$