1. **Problem Statement:** (a) Given $f(\theta) = 5 \cos \theta + \sin \theta$, express it in the form $f(\theta) = R \cos(\theta - \alpha)$, with $R>0$ and $0 \leq \alpha \leq \frac{\pi}{2}$. 2. **Find $R$ and $\alpha$:** Using the identity $R \cos(\theta - \alpha) = R(\cos \theta \cos \alpha + \sin \theta \sin \alpha)$, match coefficients: $$5 = R \cos \alpha$$ $$1 = R \sin \alpha$$ Square and add: $$R^2 = 5^2 + 1^2 = 25 + 1 = 26$$ So, $$R = \sqrt{26} \approx 5.099$$ Find $\alpha$: $$\tan \alpha = \frac{1}{5} = 0.2$$ $$\alpha = \arctan(0.2) \approx 0.197 \text{ radians}$$ 3. **Solve $5 \cos \theta + \sin \theta = 2$ for $0 \leq \theta \leq 2 \pi$:** Rewrite as: $$R \cos(\theta - \alpha) = 2$$ $$\cos(\theta - \alpha) = \frac{2}{R} = \frac{2}{5.099} \approx 0.392$$ Solutions: $$\theta - \alpha = \arccos(0.392) \text{ or } 2\pi - \arccos(0.392)$$ Calculate: $$\arccos(0.392) \approx 1.166 \text{ radians}$$ So, $$\theta_1 = 1.166 + 0.197 = 1.363$$ $$\theta_2 = 2\pi - 1.166 + 0.197 = 5.314$$ 4. **Calculate minimum value of $5 \cos 4x + \sin 4x + 15$:** Rewrite $5 \cos 4x + \sin 4x$ as $R \cos(4x - \alpha)$ where $R=\sqrt{26} \approx 5.099$ and $\alpha=0.197$ radians from previous step. Minimum of $R \cos(4x - \alpha)$ is $-R = -5.099$. Hence, Minimum value of expression = $-5.099 + 15 = 9.901$. 5. **Find smallest positive $x$ where minimum occurs:** Min occurs when: $$\cos(4x - \alpha) = -1$$ So, $$4x - \alpha = \pi$$ $$4x = \pi + 0.197 = 3.339$$ $$x = \frac{3.339}{4} = 0.835$$ 6. **Next Problem:** (a) Express $2 \sin x - 3 \cos x$ in form $R \sin(x - \alpha)$. Compare: $$2 = R \cos \alpha$$ $$-3 = -R \sin \alpha$$ (since $\sin(x - \alpha) = \sin x \cos \alpha - \cos x \sin \alpha$) So, $$2 = R \cos \alpha$$ $$3 = R \sin \alpha$$ Calculate $R$: $$R = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.606$$ Calculate $\alpha$: $$\tan \alpha = \frac{3}{2} = 1.5$$ $$\alpha = \arctan(1.5) \approx 0.983 \text{ radians}$$ (b) Greatest value of $(2 \sin x - 3 \cos x)^2$ is $R^2 = 13$. Maximum of $2 \sin x - 3 \cos x$ is $R = 3.606$. Max occurs when: $$x - \alpha = \frac{\pi}{2}$$ $$x = \frac{\pi}{2} + 0.983 = 2.554$$ (c) Solve $2 \sin x - 3 \cos x = 1$ for $0 \leq x \leq 2 \pi$. Rewrite: $$R \sin(x - \alpha) = 1$$ $$\sin(x - \alpha) = \frac{1}{R} = \frac{1}{3.606} = 0.277$$ Solutions: $$x - \alpha = \arcsin(0.277) \approx 0.281$$ or $$x - \alpha = \pi - 0.281 = 2.861$$ Calculate $x$ values: $$x_1 = 0.281 + 0.983 = 1.264$$ $$x_2 = 2.861 + 0.983 = 3.844$$ Both are within $[0, 2\pi]$. **Final Answers:** (a) $R \approx 5.099$, $\alpha \approx 0.197$ radians. (b) Solutions for $\theta$: $1.363$, $5.314$ radians. (c) Minimum value: $9.901$. (d) Smallest positive $x$ for minimum: $0.835$. (a) $R \approx 3.606$, $\alpha \approx 0.983$ radians. (b) Greatest value: $13$, smallest positive $x$ for maximum: $2.554$. (c) Solutions for $x$: $1.264$, $3.844$ radians.