1. Let's start by understanding the problem: We need to solve the equations for $0 \leq \theta \leq 2\pi$. 2. The first equation is $\sin 3\theta = \frac{\sqrt{3}}{2}$. 3. Important rule: The sine function equals $\frac{\sqrt{3}}{2}$ at angles $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ within one full circle $[0, 2\pi]$. 4. Since the equation is $\sin 3\theta = \frac{\sqrt{3}}{2}$, we set: $$3\theta = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad 3\theta = \frac{2\pi}{3} + 2k\pi$$ where $k$ is any integer because sine repeats every $2\pi$. 5. Solve for $\theta$: $$\theta = \frac{\frac{\pi}{3} + 2k\pi}{3} = \frac{\pi}{9} + \frac{2k\pi}{3}$$ $$\theta = \frac{\frac{2\pi}{3} + 2k\pi}{3} = \frac{2\pi}{9} + \frac{2k\pi}{3}$$ 6. Now find all $\theta$ values between $0$ and $2\pi$ by plugging in integer values for $k$: - For $\theta = \frac{\pi}{9} + \frac{2k\pi}{3}$, try $k=0,1,2$: - $k=0$: $\theta = \frac{\pi}{9}$ - $k=1$: $\theta = \frac{\pi}{9} + \frac{2\pi}{3} = \frac{7\pi}{9}$ - $k=2$: $\theta = \frac{\pi}{9} + \frac{4\pi}{3} = \frac{13\pi}{9}$ - For $\theta = \frac{2\pi}{9} + \frac{2k\pi}{3}$, try $k=0,1,2$: - $k=0$: $\theta = \frac{2\pi}{9}$ - $k=1$: $\theta = \frac{2\pi}{9} + \frac{2\pi}{3} = \frac{8\pi}{9}$ - $k=2$: $\theta = \frac{2\pi}{9} + \frac{4\pi}{3} = \frac{14\pi}{9}$ 7. So the solutions for part (a) are: $$\theta = \frac{\pi}{9}, \frac{7\pi}{9}, \frac{13\pi}{9}, \frac{2\pi}{9}, \frac{8\pi}{9}, \frac{14\pi}{9}$$ --- 8. Now for part (b): $\cos 3\theta - 1 = 0$ which means $\cos 3\theta = 1$. 9. Important rule: Cosine equals 1 at angles $0, 2\pi, 4\pi, ...$ i.e. $2k\pi$ for any integer $k$. 10. Set: $$3\theta = 2k\pi$$ 11. Solve for $\theta$: $$\theta = \frac{2k\pi}{3}$$ 12. Find all $\theta$ between $0$ and $2\pi$ by trying $k=0,1,2,3$: - $k=0$: $\theta=0$ - $k=1$: $\theta=\frac{2\pi}{3}$ - $k=2$: $\theta=\frac{4\pi}{3}$ - $k=3$: $\theta=2\pi$ 13. So the solutions for part (b) are: $$\theta = 0, \frac{2\pi}{3}, \frac{4\pi}{3}, 2\pi$$ These steps show how to find all angles $\theta$ that satisfy the equations within the given interval by using the periodic properties of sine and cosine functions and solving for $\theta$.