1. **Problem statement:** Given the expression $6\cos 2x - 8\sin 2x$ which can be written as $R \cos(2x + \alpha)$ where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$. Also given that $6\cos 2x - 8\sin 2x = 5$. We also have the expression $24\cos^2 x - 32\sin x \cos x$ which can be expressed in the form $a \cos 2x + b \sin 2x + c$ where $a,b,c \in \mathbb{R}$. We need to find the minimum value of $24\cos^2 x - 32\sin x \cos x$ using the above. 2. **Expressing $6\cos 2x - 8\sin 2x$ as $R \cos(2x + \alpha)$:** Recall the formula: $$R \cos(\theta + \alpha) = R \cos \theta \cos \alpha - R \sin \theta \sin \alpha$$ Comparing with $6\cos 2x - 8\sin 2x$, we get: $$6 = R \cos \alpha$$ $$-8 = -R \sin \alpha \implies 8 = R \sin \alpha$$ 3. **Find $R$ and $\alpha$:** Calculate $R$: $$R = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$ Calculate $\alpha$: $$\tan \alpha = \frac{8}{6} = \frac{4}{3}$$ Since $0 < \alpha < \frac{\pi}{2}$, $\alpha = \arctan \frac{4}{3}$. 4. **Given $6\cos 2x - 8\sin 2x = 5$, rewrite as:** $$10 \cos(2x + \alpha) = 5$$ $$\cos(2x + \alpha) = \frac{5}{10} = \frac{1}{2}$$ 5. **Express $24\cos^2 x - 32\sin x \cos x$ in terms of $\cos 2x$ and $\sin 2x$:** Recall the double angle identities: $$\cos 2x = 2\cos^2 x - 1 \implies \cos^2 x = \frac{1 + \cos 2x}{2}$$ $$\sin 2x = 2 \sin x \cos x \implies \sin x \cos x = \frac{\sin 2x}{2}$$ Substitute into the expression: $$24 \cos^2 x - 32 \sin x \cos x = 24 \times \frac{1 + \cos 2x}{2} - 32 \times \frac{\sin 2x}{2}$$ $$= 12 (1 + \cos 2x) - 16 \sin 2x$$ $$= 12 + 12 \cos 2x - 16 \sin 2x$$ So, $$a = 12, \quad b = -16, \quad c = 12$$ 6. **Find the minimum value of $24\cos^2 x - 32\sin x \cos x$:** Rewrite the variable part: $$12 \cos 2x - 16 \sin 2x$$ Express as $R \cos(2x + \beta)$: Calculate $R$: $$R = \sqrt{12^2 + (-16)^2} = \sqrt{144 + 256} = \sqrt{400} = 20$$ Calculate $\beta$: $$\tan \beta = \frac{16}{12} = \frac{4}{3}$$ Since $b$ is negative, $\beta$ is in the fourth quadrant, but for minimum and maximum values, the amplitude $R$ is sufficient. The expression becomes: $$12 + 20 \cos(2x + \beta)$$ The minimum value occurs when $\cos(2x + \beta) = -1$: $$\text{Minimum} = 12 - 20 = -8$$ **Final answers:** - $R = 10$ - $\alpha = \arctan \frac{4}{3}$ - $a = 12$, $b = -16$, $c = 12$ - Minimum value of $24\cos^2 x - 32\sin x \cos x$ is $-8$