1. **Problem statement:** Given $f(x) = 3 \cos x - 4 \sin x$ and that $f(x) = R \cos(x + \alpha)$ with $R > 0$ and $0 \leq \alpha \leq 90^\circ$, find $R$ and $\alpha$. 2. **Formula and rules:** We use the identity: $$R \cos(x + \alpha) = R \cos x \cos \alpha - R \sin x \sin \alpha$$ Matching coefficients with $3 \cos x - 4 \sin x$, we get: $$R \cos \alpha = 3$$ $$R \sin \alpha = 4$$ 3. **Find $R$:** Square and add both equations: $$R^2 \cos^2 \alpha + R^2 \sin^2 \alpha = 3^2 + 4^2$$ $$R^2 (\cos^2 \alpha + \sin^2 \alpha) = 9 + 16$$ $$R^2 = 25$$ $$R = 5$$ 4. **Find $\alpha$:** Divide the two equations: $$\tan \alpha = \frac{R \sin \alpha}{R \cos \alpha} = \frac{4}{3}$$ $$\alpha = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.1^\circ$$ 5. **Final answers for (a):** $$R = 5, \quad \alpha \approx 53.1^\circ$$