Let's solve each equation individually in the interval $[0, \pi)$.

1. $\sin(3x) = -\frac{\sqrt{3}}{2}$

We know that $\sin(\theta) = -\frac{\sqrt{3}}{2}$ when: $\theta = \frac{4\pi}{3} \quad \text{and} \quad \theta = \frac{5\pi}{3}$ Thus: $3x = \frac{4\pi}{3} \quad \text{and} \quad 3x = \frac{5\pi}{3}$ Solving for $x$: $x = \frac{4\pi}{9} \quad \text{and} \quad x = \frac{5\pi}{9}$ Therefore, for $\sin(3x) = -\frac{\sqrt{3}}{2}$: $x = \frac{4\pi}{9}, \frac{5\pi}{9}$

2. $\csc(x) = \frac{2}{\sqrt{3}}$

Since $\csc(x) = \frac{1}{\sin(x)}$, this equation is equivalent to: $\sin(x) = \frac{\sqrt{3}}{2}$ The solutions to $\sin(x) = \frac{\sqrt{3}}{2}$ in $[0, \pi)$ are: $x = \frac{\pi}{3} \quad \text{and} \quad x = \frac{2\pi}{3}$ So, for $\csc(x) = \frac{2}{\sqrt{3}}$: $x = \frac{\pi}{3}, \frac{2\pi}{3}$

3. $\cos(x) = -\frac{\sqrt{2}}{2}$

We know that $\cos(x) = -\frac{\sqrt{2}}{2}$ when: $x = \frac{3\pi}{4} \quad \text{and} \quad x = \frac{5\pi}{4}$ However, since we are only interested in the interval $[0, \pi)$, we only take: $x = \frac{3\pi}{4}$ Thus, for $\cos(x) = -\frac{\sqrt{2}}{2}$: $x = \frac{3\pi}{4}$

Final Answer

Combining all solutions, we get: $x = \frac{4\pi}{9}, \frac{5\pi}{9}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{3\pi}{4}$

Let me know if you need further clarification on any step!

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1. What are the solutions for $\sin(3x) = \frac{\sqrt{3}}{2}$ in $[0, \pi)$?
2. How does the range of $\sin(x)$ limit the solutions for $\csc(x)$?
3. What are other forms to express $-\frac{\sqrt{3}}{2}$ in trigonometry?
4. How would the solution change if we expanded the interval to $[0, 2\pi)$?
5. Can you explain how to verify these solutions on the unit circle?

Tip: Always verify trigonometric solutions by substituting back into the original equation for accuracy!