To solve for $\theta$ where $0^\circ \leq \theta < 360^\circ$ and $\sin \theta = -0.8168$, follow these steps:

Step-by-Step Solution:

1. Recognize the range and sine value:
   Since $\sin \theta$ is negative, the angle $\theta$ must be in Quadrant III or Quadrant IV.

2. Find the reference angle:
   Use the inverse sine function (arcsin) to find the reference angle $\theta_r$.
   $\theta_r = \arcsin(0.8168)$
   Using a calculator (in degree mode):
   $\theta_r \approx \arcsin(0.8168) \approx 54.02^\circ$
   This is the reference angle (always positive).

3. Determine the angles in Quadrant III and IV:

   • In Quadrant III, the angle is:
     $\theta = 180^\circ + \theta_r$
     Substituting $\theta_r = 54.02^\circ$:
     $\theta = 180^\circ + 54.02^\circ = 234.02^\circ$
   • In Quadrant IV, the angle is:
     $\theta = 360^\circ - \theta_r$
     Substituting $\theta_r = 54.02^\circ$:
     $\theta = 360^\circ - 54.02^\circ = 305.98^\circ$

4. Write the final answers:
   The solutions for $\theta$ are:
   $\theta = 234.02^\circ, 305.98^\circ$

Final Answer:

$\theta = 234.02, 305.98$

Let me know if you need further clarification or additional details!

Related Questions:

1. How do you determine the reference angle for a sine value?
2. Why is sine negative in Quadrants III and IV?
3. What is the relationship between sine and the unit circle?
4. How can you solve similar problems for cosine and tangent?
5. What are the key properties of angles in the four quadrants?

Tip: Always use the reference angle to find the angles in the appropriate quadrants based on the sign of the trigonometric function.