1. **Problem statement:** Given functions $f(x) = -2\cos x$ and $g(x) = \sin 2x$ for $-90^\circ \leq x \leq 180^\circ$, we need to: - Draw graphs of $f$ and $g$ on the same axes. - Find values of $x$ for certain conditions involving $f$ and $g$. - Find the equation of a transformed function $h(x)$. 2. **Graphing $f$ and $g$:** - $f(x) = -2\cos x$ is a cosine wave scaled by 2 and reflected vertically. - $g(x) = \sin 2x$ is a sine wave with double frequency. - Intercepts occur where the functions cross the axes. - Turning points are maxima and minima where derivatives are zero. 3. **Intercepts and turning points:** - For $f(x)$: - Intercepts: Solve $-2\cos x = 0 \Rightarrow \cos x = 0$ at $x = 90^\circ$. - Turning points: Max at $x=180^\circ$ ($f(180^\circ) = -2\cos 180^\circ = 2$), min at $x=0^\circ$ ($f(0^\circ) = -2$). - For $g(x)$: - Intercepts: Solve $\sin 2x = 0 \Rightarrow 2x = n\pi$, so $x = n\times 90^\circ$. - Turning points: Max at $x=45^\circ$ ($g(45^\circ) = 1$), min at $x=135^\circ$ ($g(135^\circ) = -1$). 4. **Find $x$ where $g(x) - f(x) = 2$:** $$\sin 2x - (-2\cos x) = 2 \Rightarrow \sin 2x + 2\cos x = 2$$ Use 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 $|\sin x + 1| \leq 2$, check values in domain $-90^\circ$ to $180^\circ$. Try $x=0^\circ$: $\cos 0^\circ = 1$, $\sin 0^\circ + 1 = 1$, product = 1 ✔ Try $x=90^\circ$: $\cos 90^\circ=0$, product=0 ✘ Try $x=30^\circ$: $\cos 30^\circ=\sqrt{3}/2 \approx 0.866$, $\sin 30^\circ +1=0.5+1=1.5$, product $\approx 1.299$ >1 ✘ Try $x=-90^\circ$: $\cos -90^\circ=0$, product=0 ✘ So solution is $x=0^\circ$. 5. **Find $x$ where $f(x) \leq g(x)$:** $$-2\cos x \leq \sin 2x$$ Rewrite: $$-2\cos x - \sin 2x \leq 0$$ Use $\sin 2x = 2\sin x \cos x$: $$-2\cos x - 2\sin x \cos x \leq 0$$ Factor: $$-2\cos x (1 + \sin x) \leq 0$$ Divide by -2 (flip inequality): $$\cos x (1 + \sin x) \geq 0$$ Analyze sign of $\cos x$ and $1 + \sin x$ over $-90^\circ$ to $180^\circ$: - $\cos x \geq 0$ on $[-90^\circ, 90^\circ]$ except near $90^\circ$. - $1 + \sin x \geq 0$ always since $\sin x \geq -1$. So $f(x) \leq g(x)$ where $\cos x (1 + \sin x) \geq 0$, approximately $x \in [-90^\circ, 90^\circ]$. 6. **Find $x$ where $f(x)$ and $g(x)$ are both increasing:** - $f(x) = -2\cos x$, derivative: $$f'(x) = -2(-\sin x) = 2\sin x$$ - $g(x) = \sin 2x$, derivative: $$g'(x) = 2\cos 2x$$ Both increasing means: $$f'(x) > 0 \Rightarrow 2\sin x > 0 \Rightarrow \sin x > 0$$ $$g'(x) > 0 \Rightarrow 2\cos 2x > 0 \Rightarrow \cos 2x > 0$$ - $\sin x > 0$ for $x \in (0^\circ, 180^\circ)$ - $\cos 2x > 0$ means $2x$ in intervals where cosine positive: $$2x \in (-90^\circ, 90^\circ) \cup (270^\circ, 450^\circ)$$ Divide by 2: $$x \in (-45^\circ, 45^\circ) \cup (135^\circ, 225^\circ)$$ Intersection with $\sin x > 0$ is: $$x \in (0^\circ, 45^\circ)$$ 7. **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: $$h(x) = 2\cos(x - 30^\circ)$$