We are tasked with finding all possible values of $\theta$ (denoted by $\emptyset$) that satisfy the following conditions:

1. $\sin \theta < 0$
2. $\sec \theta = -2$
3. The interval is $-2\pi \leq \theta \leq 2\pi$

Step 1: Use the relationship between $\sec \theta$ and $\cos \theta$

Recall that: $\sec \theta = \frac{1}{\cos \theta}$ So, if $\sec \theta = -2$, we can express this as: $\frac{1}{\cos \theta} = -2$ Solving for $\cos \theta$: $\cos \theta = -\frac{1}{2}$

Step 2: Determine the quadrants where $\cos \theta = -\frac{1}{2}$

The cosine function is negative in the second and third quadrants.

Step 3: Use reference angles

The reference angle $\theta_r$ for $\cos \theta = \frac{1}{2}$ corresponds to $\theta_r = \frac{\pi}{3}$.

Thus, in the second quadrant, the angle $\theta$ is: $\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ In the third quadrant, the angle $\theta$ is: $\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$

Step 4: Apply the condition $\sin \theta < 0$

Since sine is negative in the third and fourth quadrants, we can discard the second quadrant solution $\theta = \frac{2\pi}{3}$, as sine is positive there.

Therefore, the solution in the third quadrant, $\theta = \frac{4\pi}{3}$, is valid because both $\sin \theta < 0$ and $\cos \theta = -\frac{1}{2}$.

Step 5: Consider the periodicity of trigonometric functions

The cosine and sine functions are periodic with a period of $2\pi$, so we need to find all solutions within the interval $-2\pi \leq \theta \leq 2\pi$.

First solution: $\theta = \frac{4\pi}{3}$

Since this is already in the desired interval, it is one solution.

Second solution: $\theta = -\frac{4\pi}{3}$

We can subtract $2\pi$ from $\frac{4\pi}{3}$ to obtain another solution: $\theta = \frac{4\pi}{3} - 2\pi = -\frac{2\pi}{3}$ This solution is also valid within the interval $-2\pi \leq \theta \leq 2\pi$.

Step 6: Conclusion

The possible values of $\theta$ that satisfy the conditions $\sin \theta < 0$ and $\sec \theta = -2$ in the interval $-2\pi \leq \theta \leq 2\pi$ are: $\boxed{\theta = \frac{4\pi}{3}, -\frac{4\pi}{3}}$

Would you like more details or have any further questions?

---

Related Questions:

1. What is the general solution for $\sec \theta = -2$?
2. How is the reference angle determined for cosine and sine functions?
3. Why is sine negative in the third and fourth quadrants?
4. How does periodicity affect trigonometric function solutions?
5. How do you solve trigonometric inequalities involving sine?

Tip: Always remember to check the quadrant for both sine and cosine when working with trigonometric inequalities and identities.