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The image shows four trigonometric equations, and you're asked to solve for \( \theta \) in the range from \( 0^\circ \) to \( 360^\circ \). Let's solve each one step by step:

### (i) \( 4 \cos^2 \theta = 3 \)

1. Divide both sides by 4:
   \[
   \cos^2 \theta = \frac{3}{4}
   \]

2. Take the square root of both sides:
   \[
   \cos \theta = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}
   \]

3. So, \( \cos \theta = \frac{\sqrt{3}}{2} \) or \( \cos \theta = -\frac{\sqrt{3}}{2} \).

   From standard cosine values:
   - \( \cos \theta = \frac{\sqrt{3}}{2} \) at \( \theta = 30^\circ, 330^\circ \)
   - \( \cos \theta = -\frac{\sqrt{3}}{2} \) at \( \theta = 150^\circ, 210^\circ \)

   Therefore, the solutions for (i) are \( \theta = 30^\circ, 150^\circ, 210^\circ, 330^\circ \).

---

### (ii) \( 2 \cos^2 \theta = \cos \theta \)

1. Rearrange the equation to form a quadratic:
   \[
   2 \cos^2 \theta - \cos \theta = 0
   \]

2. Factor the equation:
   \[
   \cos \theta (2 \cos \theta - 1) = 0
   \]

3. Solve each factor:
   - \( \cos \theta = 0 \) → \( \theta = 90^\circ, 270^\circ \)
   - \( 2 \cos \theta - 1 = 0 \) → \( \cos \theta = \frac{1}{2} \)

     From standard cosine values:
     - \( \cos \theta = \frac{1}{2} \) at \( \theta = 60^\circ, 300^\circ \)

   Therefore, the solutions for (ii) are \( \theta = 60^\circ, 90^\circ, 270^\circ, 300^\circ \).

---

### (iii) \( 4 \sin \theta \cos \theta = \sin \theta \)

1. Divide both sides by \( \sin \theta \), assuming \( \sin \theta \neq 0 \):
   \[
   4 \cos \theta = 1
   \]

2. Solve for \( \cos \theta \):
   \[
   \cos \theta = \frac{1}{4}
   \]

   Using a calculator or reference for non-standard cosine values:
   - \( \theta \approx 75.52^\circ, 284.48^\circ \)

3. Additionally, consider \( \sin \theta = 0 \) (the case we ignored):
   - \( \sin \theta = 0 \) at \( \theta = 0^\circ, 180^\circ, 360^\circ \)

   Therefore, the solutions for (iii) are \( \theta = 0^\circ, 75.52^\circ, 180^\circ, 284.48^\circ, 360^\circ \).

---

### (iv) \( \cos^2 \theta - \cos \theta - 2 = 0 \)

1. This is a quadratic equation. Let \( x = \cos \theta \), so the equation becomes:
   \[
   x^2 - x - 2 = 0
   \]

2. Factor the quadratic:
   \[
   (x - 2)(x + 1) = 0
   \]

3. Solve each factor:
   - \( x - 2 = 0 \) → \( \cos \theta = 2 \) (no solution, since \( \cos \theta \) cannot be greater than 1)
   - \( x + 1 = 0 \) → \( \cos \theta = -1 \)

   From standard cosine values:
   - \( \cos \theta = -1 \) at \( \theta = 180^\circ \)

   Therefore, the solution for (iv) is \( \theta = 180^\circ \).

---

### Summary of Solutions
1. (i) \( \theta = 30^\circ, 150^\circ, 210^\circ, 330^\circ \)
2. (ii) \( \theta = 60^\circ, 90^\circ, 270^\circ, 300^\circ \)
3. (iii) \( \theta = 0^\circ, 75.52^\circ, 180^\circ, 284.48^\circ, 360^\circ \)
4. (iv) \( \theta = 180^\circ \)

Would you like more details or clarification on any of the steps?

Here are 5 related questions for further practice:
1. How do you derive the general solutions for trigonometric equations?
2. What is the method for solving higher-order trigonometric equations?
3. How do the unit circle and standard angles assist in solving trigonometric equations?
4. What are the limitations on the range of \( \sin \theta \) and \( \cos \theta \)?
5. Can you explain how the quadratic factoring method applies to trigonometric functions?

**Tip:** When solving trigonometric equations, always check if you can factor expressions or use known identities to simplify the process.

Solve the following trigonometric equations for θ in the range from 0° to 360°:
(i) 4cos²θ = 3
(ii) 2cos²θ = cosθ
(iii) 4sinθcosθ = sinθ
(iv) cos²θ - cosθ - 2 = 0

This problem involves solving four trigonometric equations, each requiring the application of identities, quadratic equations, and known sine and cosine values. Step-by-step solutions for each equation are provided. Suitable for students in Grades 10-12.

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