To find the angle $\theta$ in trigonometric equations like $\cos \theta = -\frac{\sqrt{3}}{2}$, it helps to understand the properties of cosine and the specific values it takes on the unit circle. Here’s a step-by-step explanation:

1. Understanding the Value $-\frac{\sqrt{3}}{2}$:

   • For cosine, $\frac{\sqrt{3}}{2}$ is a known value corresponding to $30^\circ$ (or $\pi/6$ radians) and $150^\circ$ (or $5\pi/6$ radians).
   • However, since our value is negative, it must correspond to an angle in either the second or third quadrant, where cosine values are negative.

2. Identifying the Quadrant:

   • The cosine function is negative in the second and third quadrants. Knowing this, we look at the angles corresponding to $\cos \theta = \frac{\sqrt{3}}{2}$, which are $30^\circ$ from the x-axis in either the second or third quadrant.

3. Finding the Angles:

   • In the second quadrant, the angle would be $180^\circ - 30^\circ = 150^\circ$.
   • In the third quadrant, the angle would be $180^\circ + 30^\circ = 210^\circ$.

4. Answer:

   • The angles that satisfy $\cos \theta = -\frac{\sqrt{3}}{2}$ are $\theta = 150^\circ$ or $\theta = 210^\circ$.

So the trick here is recognizing the reference angle (here $30^\circ$), understanding which quadrants make the cosine negative, and adjusting the angle accordingly based on the quadrant.

Would you like more details on this process or how to handle other trigonometric functions?

---

1. What if the equation was $\sin \theta = \pm \frac{\sqrt{3}}{2}$?
2. How does the process change if we work in radians instead of degrees?
3. What would be the steps if the problem was $\tan \theta = -\sqrt{3}$?
4. How can we use reference triangles for different trigonometric values?
5. Can we generalize this process for any value of $\cos \theta$?

Tip: Memorizing the key angles and values for $\sin$, $\cos$, and $\tan$ on the unit circle simplifies solving these types of problems.