1. **State the problem:** Express $\sqrt{15} \sin(2x) + \sqrt{5} \cos(2x)$ in the form $R \sin(2x + \alpha)$. 2. **Formula and rules:** We use the identity: $$R \sin(\theta + \alpha) = R \sin \theta \cos \alpha + R \cos \theta \sin \alpha$$ Here, $\theta = 2x$. 3. **Match coefficients:** From the problem, $$\sqrt{15} \sin(2x) + \sqrt{5} \cos(2x) = R \sin(2x) \cos \alpha + R \cos(2x) \sin \alpha$$ Equate coefficients: $$R \cos \alpha = \sqrt{15}$$ $$R \sin \alpha = \sqrt{5}$$ 4. **Find $R$:** $$R = \sqrt{(R \cos \alpha)^2 + (R \sin \alpha)^2} = \sqrt{(\sqrt{15})^2 + (\sqrt{5})^2} = \sqrt{15 + 5} = \sqrt{20} = 2\sqrt{5}$$ 5. **Find $\alpha$:** $$\tan \alpha = \frac{R \sin \alpha}{R \cos \alpha} = \frac{\sqrt{5}}{\sqrt{15}} = \frac{\sqrt{5}}{\sqrt{15}} = \frac{1}{\sqrt{3}}$$ So, $$\alpha = \arctan \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6}$$ 6. **Rewrite expression:** $$\sqrt{15} \sin(2x) + \sqrt{5} \cos(2x) = 2\sqrt{5} \sin \left( 2x + \frac{\pi}{6} \right)$$ --- 7. **Part b:** Given $$f(x) = \frac{2}{5 + \sqrt{15} \sin(2x) + \sqrt{5} \cos(2x)}$$ Using part a, $$f(x) = \frac{2}{5 + 2\sqrt{5} \sin \left( 2x + \frac{\pi}{6} \right)}$$ 8. **Maximum value of $f(x)$:** The denominator is minimized when $\sin \left( 2x + \frac{\pi}{6} \right) = -1$ (minimum sine value), but since it is added, the maximum of $f(x)$ occurs when denominator is minimum. Minimum denominator: $$5 + 2\sqrt{5} \times (-1) = 5 - 2\sqrt{5}$$ Maximum $f(x)$: $$f_{max} = \frac{2}{5 - 2\sqrt{5}}$$ 9. **Rationalize denominator:** Multiply numerator and denominator by the conjugate $5 + 2\sqrt{5}$: $$f_{max} = \frac{2(5 + 2\sqrt{5})}{(5 - 2\sqrt{5})(5 + 2\sqrt{5})} = \frac{2(5 + 2\sqrt{5})}{25 - (2\sqrt{5})^2} = \frac{2(5 + 2\sqrt{5})}{25 - 4 \times 5} = \frac{2(5 + 2\sqrt{5})}{25 - 20} = \frac{2(5 + 2\sqrt{5})}{5}$$ Simplify: $$f_{max} = \frac{2}{5} \times (5 + 2\sqrt{5}) = 2 + \frac{4}{5} \sqrt{5}$$ So, $$p = 2, \quad q = \frac{4}{5}$$ 10. **Smallest $x$ for maximum:** Maximum occurs when $$\sin \left( 2x + \frac{\pi}{6} \right) = -1$$ This happens at $$2x + \frac{\pi}{6} = \frac{3\pi}{2} + 2k\pi, \quad k \in \mathbb{Z}$$ For smallest $x$, take $k=0$: $$2x = \frac{3\pi}{2} - \frac{\pi}{6} = \frac{9\pi}{6} - \frac{\pi}{6} = \frac{8\pi}{6} = \frac{4\pi}{3}$$ Therefore, $$x = \frac{2\pi}{3}$$ --- **Final answers:** - a) $\sqrt{15} \sin(2x) + \sqrt{5} \cos(2x) = 2\sqrt{5} \sin \left( 2x + \frac{\pi}{6} \right)$ - b i) $\max f(x) = 2 + \frac{4}{5} \sqrt{5}$ - b ii) smallest $x$ for max is $x = \frac{2\pi}{3}$