Let's solve the given trigonometric equations for $x \in [0, 2\pi]$ one by one:

a) $\csc x = -2$

We know the reciprocal relationship:
$\csc x = \frac{1}{\sin x}$
Thus, $\sin x = -\frac{1}{2}$.

The sine function is negative in the third and fourth quadrants. Therefore, the solutions are:
$x = \pi + \arcsin\left(\frac{1}{2}\right) \quad \text{and} \quad x = 2\pi - \arcsin\left(\frac{1}{2}\right)$

Since $\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$, the solutions are:
$x = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \quad \text{and} \quad x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$

Final solutions for (a): $x = \frac{7\pi}{6}, \frac{11\pi}{6}$

b) $\csc\left(x - \frac{\pi}{4}\right) = -2$

Using the same reciprocal relationship,
$\sin\left(x - \frac{\pi}{4}\right) = -\frac{1}{2}.$

The sine function is negative in the third and fourth quadrants. Thus,
$x - \frac{\pi}{4} = \pi + \arcsin\left(\frac{1}{2}\right) \quad \text{and} \quad x - \frac{\pi}{4} = 2\pi - \arcsin\left(\frac{1}{2}\right).$

As $\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$, we substitute:
$x - \frac{\pi}{4} = \pi + \frac{\pi}{6} \quad \text{and} \quad x - \frac{\pi}{4} = 2\pi - \frac{\pi}{6}.$

Solving for $x$:

1. $x = \pi + \frac{\pi}{6} + \frac{\pi}{4} = \frac{6\pi}{6} + \frac{1\pi}{6} + \frac{3\pi}{12} = \frac{17\pi}{12}$
2. $x = 2\pi - \frac{\pi}{6} + \frac{\pi}{4} = \frac{12\pi}{6} - \frac{1\pi}{6} + \frac{3\pi}{12} = \frac{23\pi}{12}$.

Final solutions for (b): $x = \frac{17\pi}{12}, \frac{23\pi}{12}$

c) $3\sec x = 2\sqrt{3}$

Dividing through by 3:
$\sec x = \frac{2\sqrt{3}}{3}.$

Taking the reciprocal relationship,
$\cos x = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}.$

The cosine function is positive in the first and fourth quadrants. Thus,
$x = \arccos\left(\frac{\sqrt{3}}{2}\right) \quad \text{and} \quad x = 2\pi - \arccos\left(\frac{\sqrt{3}}{2}\right).$

Since $\arccos\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$, the solutions are:
$x = \frac{\pi}{6} \quad \text{and} \quad x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}.$

Final solutions for (c): $x = \frac{\pi}{6}, \frac{11\pi}{6}$

Summary of All Solutions:

• (a): $x = \frac{7\pi}{6}, \frac{11\pi}{6}$
• (b): $x = \frac{17\pi}{12}, \frac{23\pi}{12}$
• (c): $x = \frac{\pi}{6}, \frac{11\pi}{6}$

Would you like a detailed explanation of any part? Here are some follow-up questions:

1. How do we derive the reciprocal relationships for trigonometric functions?
2. What are the general solutions for $\sin x = k$ and $\cos x = k$?
3. Why is $\arcsin k$ only defined in $[-\frac{\pi}{2}, \frac{\pi}{2}]$?
4. Can you explain the significance of the unit circle in solving these equations?
5. How would the solutions differ if the range for $x$ were not $[0, 2\pi]$?

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