1. **State the problem:** We want to express the equation $$3 \cos x + 2\sqrt{10} \sin x = R \cos(x - \alpha)$$ and find the value of $R$ where $R > 0$. 2. **Formula used:** The expression $a \cos x + b \sin x$ can be rewritten as $R \cos(x - \alpha)$ where $$R = \sqrt{a^2 + b^2}$$ and $\alpha$ is an angle such that $$\cos \alpha = \frac{a}{R}, \quad \sin \alpha = \frac{b}{R}$$ 3. **Identify coefficients:** Here, $a = 3$ and $b = 2\sqrt{10}$. 4. **Calculate $R$:** $$R = \sqrt{3^2 + (2\sqrt{10})^2} = \sqrt{9 + 4 \times 10} = \sqrt{9 + 40} = \sqrt{49}$$ 5. **Simplify:** $$R = 7$$ 6. **Conclusion:** The value of $R$ is $7$.