To solve the equation $\cos \theta = -\frac{\sqrt{3}}{2}$ in the interval $0^\circ \leq \theta < 360^\circ$, we proceed as follows:

Step 1: Identify the reference angle

The cosine of an angle equals $\frac{\sqrt{3}}{2}$ for the angles $30^\circ$ and $330^\circ$, but because we have a negative value, we need to consider the quadrants where cosine is negative.

Step 2: Determine the quadrants

Cosine is negative in the second and third quadrants.

Step 3: Find the angles in the relevant quadrants

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

Final Solution:

The solutions for $\cos \theta = -\frac{\sqrt{3}}{2}$ in the interval $0^\circ \leq \theta < 360^\circ$ are: $\theta = 150^\circ \quad \text{and} \quad \theta = 210^\circ.$

Would you like further details on how to interpret this result, or do you have any questions?

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Here are 5 related questions for further practice:

1. Solve $\sin \theta = \frac{1}{2}$ for $0^\circ \leq \theta < 360^\circ$.
2. Find the solutions to $\cos \theta = \frac{1}{2}$ in the interval $0^\circ \leq \theta < 360^\circ$.
3. What are the solutions for $\tan \theta = 1$ in $0^\circ \leq \theta < 360^\circ$?
4. Solve $\cos \theta = -\frac{1}{2}$ for $0^\circ \leq \theta < 360^\circ$.
5. Find the solutions for $\sin \theta = -\frac{\sqrt{3}}{2}$ in $0^\circ \leq \theta < 360^\circ$.

Tip: Always remember that cosine is negative in the second and third quadrants, and sine is negative in the third and fourth quadrants!