Solve this equation if 0°  greater than/equal to theta < 360° for cos theta = -square root 3/2

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:
\[
\theta = 150^\circ \quad \text{and} \quad \theta = 210^\circ.
\]

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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!

Solve this equation if 0° greater than/equal to theta < 360° for cos theta = -√3/2

Learn how to solve the trigonometric equation cos(θ) = -√3/2 in the interval 0° ≤ θ < 360°. This solution covers key trigonometry concepts like reference angles and the cosine function, identifying solutions in different quadrants. Ideal for high school students in Grades 9-12.

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