We are solving the equation:
$\sec(3\theta) - 2 = 0$
on the interval $[0, 2\pi)$.

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Step-by-Step Solution:

1. Isolate the secant term: $\sec(3\theta) = 2$

2. Convert to cosine: Recall that $\sec(x) = \frac{1}{\cos(x)}$. Therefore: $\frac{1}{\cos(3\theta)} = 2$ Multiplying through by $\cos(3\theta)$ and solving for $\cos(3\theta)$: $\cos(3\theta) = \frac{1}{2}$

3. Find the principal solutions: The cosine function equals $\frac{1}{2}$ at angles: 3\theta = \frac{\pi}{3}, \frac{5\pi}{3} \quad \text{(in the interval [0, 2\pi))}.

4. Solve for $\theta$: Divide each angle by 3 to isolate $\theta$: $\theta = \frac{\pi}{9}, \frac{5\pi}{9}.$

5. Check for periodic solutions within $[0, 2\pi)$: The cosine function is periodic with period $2\pi$, and since $3\theta$ cycles three times faster, we check for additional solutions by adding $2\pi$ to each principal solution (scaled by $1/3$): $\theta = \frac{\pi}{9}, \frac{5\pi}{9}, \frac{\pi}{9} + \frac{2\pi}{3}, \frac{5\pi}{9} + \frac{2\pi}{3}, \frac{\pi}{9} + \frac{4\pi}{3}, \frac{5\pi}{9} + \frac{4\pi}{3}.$

   Simplifying: $\theta = \frac{\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}, \frac{11\pi}{9}, \frac{13\pi}{9}, \frac{17\pi}{9}.$

6. Ensure the solutions are in $[0, 2\pi)$: All the above solutions are within the interval.

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Final Answer:

The solutions are: $\frac{\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}, \frac{11\pi}{9}, \frac{13\pi}{9}, \frac{17\pi}{9}.$

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Do you need more details or clarification?

Related Questions:

1. How do you derive the general solution for trigonometric equations?
2. What is the relationship between secant and cosine?
3. Why do we divide by 3 when solving for $\theta$ in this equation?
4. How can you graphically verify these solutions?
5. What is the importance of specifying the interval $[0, 2\pi)$?

Tip:

For periodic trigonometric equations, always account for all possible solutions within the given interval by considering the periodicity of the function.