To solve for all angles $\theta$ where $0^\circ \leq \theta < 360^\circ$ and $\cos(\theta) = -\frac{\sqrt{3}}{2}$, we follow these steps:

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Step 1: Identify the reference angle

The cosine of an angle is negative in the second and third quadrants. First, we find the reference angle, which is the acute angle in the first quadrant where the cosine value equals $\frac{\sqrt{3}}{2}$.

From trigonometric values: $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = 30^\circ$ Thus, the reference angle is $30^\circ$.

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Step 2: Determine the quadrants

Since $\cos(\theta)$ is negative, $\theta$ lies in:

• Quadrant II ($180^\circ - \text{reference angle}$)
• Quadrant III ($180^\circ + \text{reference angle}$)

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Step 3: Calculate the solutions

1. In Quadrant II: $\theta = 180^\circ - 30^\circ = 150^\circ$

2. In Quadrant III: $\theta = 180^\circ + 30^\circ = 210^\circ$

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Final Answer:

The angles that satisfy $\cos(\theta) = -\frac{\sqrt{3}}{2}$ in the interval $0^\circ \leq \theta < 360^\circ$ are: $\boxed{\theta = 150^\circ \text{ and } 210^\circ}$

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Would you like further details on these calculations or a visual representation of this on the unit circle?

Related Questions:

1. What are the general solutions for $\cos(\theta) = -\frac{\sqrt{3}}{2}$ in any interval?
2. How does the reference angle relate to trigonometric function values?
3. Can you derive the cosine of $30^\circ$ geometrically?
4. Why is cosine negative in Quadrants II and III?
5. How would the solution change if the range were $0 \leq \theta \leq 2\pi$ radians instead?

Tip:

For cosine values, remember that the sign depends on the x-coordinate of the unit circle, positive in Quadrants I and IV, and negative in Quadrants II and III.