1. **Problem (i): Solve for $0 \leq \theta < 360^\circ$ the equation $9 \sin(\theta + 60^\circ) = 4$.** 2. **Step 1: Isolate the sine function.** $$\sin(\theta + 60^\circ) = \frac{4}{9}$$ 3. **Step 2: Use the inverse sine function to find the principal value.** $$\theta + 60^\circ = \sin^{-1}\left(\frac{4}{9}\right)$$ Calculate $\sin^{-1}(4/9) \approx 26.3^\circ$. 4. **Step 3: Find the general solutions for $\theta + 60^\circ$.** Since sine is positive, solutions in $0^\circ$ to $360^\circ$ are: $$\theta + 60^\circ = 26.3^\circ \quad \text{or} \quad 180^\circ - 26.3^\circ = 153.7^\circ$$ 5. **Step 4: Solve for $\theta$.** $$\theta = 26.3^\circ - 60^\circ = -33.7^\circ$$ $$\theta = 153.7^\circ - 60^\circ = 93.7^\circ$$ 6. **Step 5: Adjust $\theta$ to be within $0^\circ \leq \theta < 360^\circ$.** Add $360^\circ$ to $-33.7^\circ$: $$\theta = 360^\circ - 33.7^\circ = 326.3^\circ$$ 7. **Final answers for (i):** $$\boxed{\theta = 93.7^\circ, 326.3^\circ}$$ --- 8. **Problem (ii): Solve for $-\pi \leq x < \pi$ the equation $2 \tan x - 3 \sin x = 0$.** 9. **Step 1: Rewrite the equation.** $$2 \tan x = 3 \sin x$$ 10. **Step 2: Express $\tan x$ as $\frac{\sin x}{\cos x}$.** $$2 \frac{\sin x}{\cos x} = 3 \sin x$$ 11. **Step 3: Multiply both sides by $\cos x$ (noting $\cos x \neq 0$).** $$2 \sin x = 3 \sin x \cos x$$ 12. **Step 4: Rearrange the equation.** $$2 \sin x - 3 \sin x \cos x = 0$$ $$\sin x (2 - 3 \cos x) = 0$$ 13. **Step 5: Solve each factor separately.** - Case 1: $\sin x = 0$ $$x = 0, \pm \pi$$ Within $-\pi \leq x < \pi$, solutions are $x = -\pi, 0$. - Case 2: $2 - 3 \cos x = 0$ $$3 \cos x = 2$$ $$\cos x = \frac{2}{3}$$ 14. **Step 6: Find $x$ values for $\cos x = \frac{2}{3}$.** $$x = \pm \cos^{-1}\left(\frac{2}{3}\right)$$ Calculate $\cos^{-1}(2/3) \approx 0.8411$ radians. 15. **Step 7: List all solutions in the interval $-\pi \leq x < \pi$.** $$x = -0.84, 0, 0.84, -\pi$$ 16. **Final answers for (ii) rounded to 2 decimal places:** $$\boxed{x = -3.14, -0.84, 0, 0.84}$$