1. **State the problem:** We need to sketch the graph of $y = |\cos(\frac{x}{4})|$ for $0 \leq x \leq 8\pi$ and solve the equation $|\cos(\frac{x}{4})| = \frac{1}{2}$ in the same interval. 2. **Graph sketching:** The function $y = |\cos(\frac{x}{4})|$ takes the cosine function with input scaled by $\frac{1}{4}$ and reflects all negative values above the x-axis. - The period of $\cos(\theta)$ is $2\pi$, so the period of $\cos(\frac{x}{4})$ is $8\pi$ because: $$\text{Period} = \frac{2\pi}{\frac{1}{4}} = 8\pi$$ - Within $0 \leq x \leq 8\pi$, the graph completes exactly one full period. - The graph peaks at $y=1$ when $\cos(\frac{x}{4})=\pm 1$, which happens at $x=0, 4\pi, 8\pi$. - The zeros of $|\cos(\frac{x}{4})|$ occur where $\cos(\frac{x}{4})=0$, i.e., at $x=2\pi$ and $6\pi$. 3. **Solving $|\cos(\frac{x}{4})| = \frac{1}{2}$:** - This means: $$\cos\left(\frac{x}{4}\right) = \pm \frac{1}{2}$$ - Recall that $\cos(\theta) = \frac{1}{2}$ at $\theta = \pm \frac{\pi}{3} + 2k\pi$ and $\cos(\theta) = -\frac{1}{2}$ at $\theta = \pm \frac{2\pi}{3} + 2k\pi$ for integers $k$. - Let $\theta = \frac{x}{4}$, so: For $\cos(\theta) = \frac{1}{2}$: $$\theta = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad \theta = -\frac{\pi}{3} + 2k\pi$$ For $\cos(\theta) = -\frac{1}{2}$: $$\theta = \frac{2\pi}{3} + 2k\pi \quad \text{or} \quad \theta = -\frac{2\pi}{3} + 2k\pi$$ - Since $0 \leq x \leq 8\pi$, then $0 \leq \theta = \frac{x}{4} \leq 2\pi$. - Find all $\theta$ in $[0, 2\pi]$ satisfying these: For $\cos(\theta) = \frac{1}{2}$: - $\theta = \frac{\pi}{3}$ - $\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$ For $\cos(\theta) = -\frac{1}{2}$: - $\theta = \frac{2\pi}{3}$ - $\theta = 2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}$ - Convert back to $x$ by multiplying by 4: $$x = 4\theta$$ So the solutions are: $$x = 4 \times \frac{\pi}{3} = \frac{4\pi}{3}$$ $$x = 4 \times \frac{5\pi}{3} = \frac{20\pi}{3}$$ $$x = 4 \times \frac{2\pi}{3} = \frac{8\pi}{3}$$ $$x = 4 \times \frac{4\pi}{3} = \frac{16\pi}{3}$$ - All these values lie within $0 \leq x \leq 8\pi$ since $8\pi = \frac{24\pi}{3}$. **Final answers:** - The graph is a cosine wave with period $8\pi$ reflected above the x-axis, peaks at $x=0,4\pi,8\pi$, zeros at $x=2\pi,6\pi$. - The solutions to $|\cos(\frac{x}{4})| = \frac{1}{2}$ in $[0,8\pi]$ are: $$x = \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{16\pi}{3}, \frac{20\pi}{3}$$