Solve : (i)` cos 3theta + 8 cos^(3) theta = 0`
(ii) `"tan" (pi cot x) = cot (pi tan x)`
(iii) `4 cos^(2) x sin x-2 "sin"^(2) x = 3 sin x`

Answer

Let's solve the given trigonometric equations step by step.

### (i) Solve: \( \cos 3\theta + 8 \cos^3 \theta = 0 \)

1. **Rewrite \( \cos 3\theta \)**:
\[ \cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta \]
Substitute this into the equation:
\[ 4 \cos^3 \theta - 3 \cos \theta + 8 \cos^3 \theta = 0 \]

2. **Combine like terms**:
\[ 12 \cos^3 \theta - 3 \cos \theta = 0 \]

3. **Factor out \( 3 \cos \theta \)**:
\[ 3 \cos \theta (4 \cos^2 \theta - 1) = 0 \]

4. **Set each factor to zero**:
- Case A: \( 3 \cos \theta = 0 \)
- Case B: \( 4 \cos^2 \theta - 1 = 0 \)

5. **Solve Case A**:
\[ \cos \theta = 0 \implies \theta = \frac{(2n + 1) \pi}{2}, \quad n \in \mathbb{Z} \]

6. **Solve Case B**:
\[ 4 \cos^2 \theta = 1 \implies \cos^2 \theta = \frac{1}{4} \implies \cos \theta = \pm \frac{1}{2} \]
- \( \cos \theta = \frac{1}{2} \implies \theta = 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z} \)
- \( \cos \theta = -\frac{1}{2} \implies \theta = 2n\pi \pm \frac{2\pi}{3}, \quad n \in \mathbb{Z} \)

### Final Solutions for (i):
\[ \theta = \frac{(2n + 1) \pi}{2}, \quad \theta = 2n\pi \pm \frac{\pi}{3}, \quad \theta = 2n\pi \pm \frac{2\pi}{3}, \quad n \in \mathbb{Z} \]