We are given that:

$\log_2(\sin(\theta)) = -2$

To find the value of $\log_2(\csc(\theta))$, let's proceed step by step:

Step 1: Solve for $\sin(\theta)$

We can rewrite the logarithmic equation:

$\log_2(\sin(\theta)) = -2 \implies \sin(\theta) = 2^{-2} = \frac{1}{4}$

Step 2: Express $\csc(\theta)$

Recall that $\csc(\theta) = \frac{1}{\sin(\theta)}$, so:

$\csc(\theta) = \frac{1}{\frac{1}{4}} = 4$

Step 3: Find $\log_2(\csc(\theta))$

Now, we find $\log_2(\csc(\theta))$:

$\log_2(\csc(\theta)) = \log_2(4)$

Since $4 = 2^2$, we have:

$\log_2(4) = 2$

Thus, the value of $\log_2(\csc(\theta))$ is:

$\boxed{2}$

Would you like a more detailed explanation of any step? Here are some related questions to expand your understanding:

1. What is the relationship between $\sin(\theta)$ and $\csc(\theta)$?
2. How do you convert between logarithmic and exponential forms?
3. What are the properties of logarithms that help in solving equations like this?
4. What is the general approach to solving $\log_b(x) = c$?
5. How can we use identities involving trigonometric functions and logarithms?

Tip: Always remember that the cosecant function is the reciprocal of the sine function, which often helps when switching between the two!