1. **Problem statement:** Express $3 \cos x + 3 \sin x$ in the form $R \cos(x - a)$ where $R > 0$ and $0 < a < \frac{\pi}{2}$. 2. **Formula and rules:** We use the identity: $$R \cos(x - a) = R \cos x \cos a + R \sin x \sin a$$ Matching coefficients with $3 \cos x + 3 \sin x$, we get: $$R \cos a = 3$$ $$R \sin a = 3$$ 3. **Find $R$:** Square and add both equations: $$R^2 \cos^2 a + R^2 \sin^2 a = 3^2 + 3^2$$ $$R^2 (\cos^2 a + \sin^2 a) = 9 + 9$$ $$R^2 = 18$$ $$R = \sqrt{18} = 3\sqrt{2}$$ 4. **Find $a$:** Divide the two equations: $$\frac{R \sin a}{R \cos a} = \frac{3}{3}$$ $$\tan a = 1$$ Since $0 < a < \frac{\pi}{2}$, we have: $$a = \frac{\pi}{4}$$ 5. **Final expression:** $$3 \cos x + 3 \sin x = 3\sqrt{2} \cos \left(x - \frac{\pi}{4}\right)$$ --- 6. **Problem statement:** Given $T(x) = \frac{8}{3 \cos x + 3 \sin x}$, (a) Determine a value of $x$ for which $T(x)$ is not defined. 7. **Solution:** $T(x)$ is undefined when the denominator is zero: $$3 \cos x + 3 \sin x = 0$$ Divide both sides by 3: $$\cancel{3} \cos x + \cancel{3} \sin x = 0 \implies \cos x + \sin x = 0$$ Rearranged: $$\sin x = -\cos x$$ Divide both sides by $\cos x$ (assuming $\cos x \neq 0$): $$\tan x = -1$$ General solution: $$x = -\frac{\pi}{4} + n\pi, \quad n \in \mathbb{Z}$$ One value in $[0, 2\pi)$ is: $$x = \frac{3\pi}{4}$$ --- 8. **Problem statement:** (b) Find the smallest positive $x$ satisfying: $$T(3x) = \frac{8}{9} \sqrt{6}$$ 9. **Solution:** Recall: $$T(3x) = \frac{8}{3 \cos(3x) + 3 \sin(3x)}$$ Set equal to given value: $$\frac{8}{3 \cos(3x) + 3 \sin(3x)} = \frac{8}{9} \sqrt{6}$$ Divide both sides by 8: $$\frac{1}{3 \cos(3x) + 3 \sin(3x)} = \frac{\sqrt{6}}{9}$$ Invert both sides: $$3 \cos(3x) + 3 \sin(3x) = \frac{9}{\sqrt{6}}$$ Divide both sides by 3: $$\cos(3x) + \sin(3x) = \frac{3}{\sqrt{6}}$$ Simplify right side: $$\frac{3}{\sqrt{6}} = \frac{3 \sqrt{6}}{6} = \frac{\sqrt{6}}{2}$$ 10. **Rewrite left side using previous form:** From step 1-5, we know: $$3 \cos y + 3 \sin y = 3\sqrt{2} \cos \left(y - \frac{\pi}{4}\right)$$ Divide by 3: $$\cos y + \sin y = \sqrt{2} \cos \left(y - \frac{\pi}{4}\right)$$ Apply to $y = 3x$: $$\sqrt{2} \cos \left(3x - \frac{\pi}{4}\right) = \frac{\sqrt{6}}{2}$$ Divide both sides by $\sqrt{2}$: $$\cos \left(3x - \frac{\pi}{4}\right) = \frac{\sqrt{6}}{2 \sqrt{2}} = \frac{\sqrt{3}}{2}$$ 11. **Solve for $3x - \frac{\pi}{4}$:** $$\cos \theta = \frac{\sqrt{3}}{2}$$ Solutions in $[0, 2\pi)$: $$\theta = \pm \frac{\pi}{6} + 2n\pi$$ So: $$3x - \frac{\pi}{4} = \pm \frac{\pi}{6} + 2n\pi$$ 12. **Find $x$:** For smallest positive $x$, take $n=0$ and positive angle: $$3x = \frac{\pi}{4} + \frac{\pi}{6} = \frac{3\pi}{12} + \frac{2\pi}{12} = \frac{5\pi}{12}$$ $$x = \frac{5\pi}{36}$$ 13. **Final answers:** - (i) $3 \cos x + 3 \sin x = 3\sqrt{2} \cos \left(x - \frac{\pi}{4}\right)$ - (ii.a) $T(x)$ undefined at $x = \frac{3\pi}{4}$ - (ii.b) Smallest positive $x$ with $T(3x) = \frac{8}{9} \sqrt{6}$ is $x = \frac{5\pi}{36}$