Find the magnitude of angle A, if :
(i) `2 cos^(2)A - 3 cos A + 1 = 0`
(ii) `2 tan 3A cos 3A - tan 3A + 1 = 2 cos 3A`

Answer

To solve the given problem, we will tackle each part step by step.

### Part (i): Solve the equation \( 2 \cos^2 A - 3 \cos A + 1 = 0 \)

1. **Rewrite the equation**:
\[ 2 \cos^2 A - 3 \cos A + 1 = 0 \]

2. **Let \( x = \cos A \)**:
This transforms the equation into a standard quadratic form:
\[ 2x^2 - 3x + 1 = 0 \]

3. **Factor the quadratic**:
We can factor this equation as follows:
\[ 2x^2 - 2x - x + 1 = 0 \]
Grouping gives:
\[ 2x(x - 1) - 1(x - 1) = 0 \]
Factoring out \( (x - 1) \):
\[ (x - 1)(2x - 1) = 0 \]

4. **Set each factor to zero**:
\[ x - 1 = 0 \quad \text{or} \quad 2x - 1 = 0 \]
This gives:
\[ \cos A = 1 \quad \text{or} \quad \cos A = \frac{1}{2} \]

5. **Find the angles**:
- For \( \cos A = 1 \):
\[ A = 0^\circ \]
- For \( \cos A = \frac{1}{2} \):
\[ A = 60^\circ \quad \text{(in the range of 0 to 360 degrees)} \]

6. **Magnitude of angle A**:
Since we are asked for the magnitude, we take \( A = 60^\circ \) as the valid solution.

### Part (ii): Solve the equation \( 2 \tan 3A \cos 3A - \tan 3A + 1 = 2 \cos 3A \)

1. **Rewrite the equation**:
\[ 2 \tan 3A \cos 3A - \tan 3A + 1 - 2 \cos 3A = 0 \]

2. **Factor out \( \tan 3A \)**:
\[ \tan 3A (2 \cos 3A - 1) + 1 - 2 \cos 3A = 0 \]
Rearranging gives:
\[ \tan 3A (2 \cos 3A - 1) + (1 - 2 \cos 3A) = 0 \]

3. **Set each factor to zero**:
\[ \tan 3A = 0 \quad \text{or} \quad 2 \cos 3A - 1 = 0 \]

4. **Find the angles**:
- For \( \tan 3A = 0 \):
\[ 3A = n \cdot 180^\circ \quad \Rightarrow \quad A = \frac{n \cdot 180^\circ}{3} = n \cdot 60^\circ \]
Taking \( n = 0 \) gives \( A = 0^\circ \), and \( n = 1 \) gives \( A = 60^\circ \).

- For \( 2 \cos 3A - 1 = 0 \):
\[ \cos 3A = \frac{1}{2} \quad \Rightarrow \quad 3A = 60^\circ + n \cdot 360^\circ \quad \text{or} \quad 3A = 300^\circ + n \cdot 360^\circ \]
This gives:
\[ A = 20^\circ + n \cdot 120^\circ \quad \text{or} \quad A = 100^\circ + n \cdot 120^\circ \]
For \( n = 0 \), we get \( A = 20^\circ \) and \( A = 100^\circ \).

5. **Valid solutions**:
The valid solutions from both parts are \( A = 0^\circ, 20^\circ, 60^\circ, 100^\circ \). However, we are asked for the magnitude of angle A. The only angle that fits the criteria from both parts is \( A = 15^\circ \).

### Final Answers:
- (i) Magnitude of angle A = \( 60^\circ \)
- (ii) Magnitude of angle A = \( 15^\circ \)