If `4 cos^(2) x = 3 and x` is an acute angle find the value of :
(i) x
(ii) `cos^(2)x + cot^(2)x`
(iii) cos 3x
(iv) sin 2x

Answer

To solve the problem step by step, we will follow the instructions provided in the question.

### Given:
\[ 4 \cos^2 x = 3 \]
where \( x \) is an acute angle.

### Step 1: Solve for \( \cos^2 x \)
We start by isolating \( \cos^2 x \):
\[ \cos^2 x = \frac{3}{4} \]

### Step 2: Find \( \cos x \)
Next, we take the square root of both sides to find \( \cos x \):
\[ \cos x = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]
Since \( x \) is an acute angle, we take the positive root.

### Step 3: Determine the angle \( x \)
We know that:
\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \]
Thus, we can conclude:
\[ x = 30^\circ \]

### Step 4: Calculate \( \cos^2 x + \cot^2 x \)
First, we need to find \( \cot x \):
\[ \cot x = \frac{\cos x}{\sin x} \]
To find \( \sin x \), we use the identity \( \sin^2 x + \cos^2 x = 1 \):
\[ \sin^2 x = 1 - \cos^2 x = 1 - \frac{3}{4} = \frac{1}{4} \]
Thus,
\[ \sin x = \sqrt{\frac{1}{4}} = \frac{1}{2} \]
Now we can find \( \cot x \):
\[ \cot x = \frac{\cos x}{\sin x} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \]
Now, we calculate \( \cot^2 x \):
\[ \cot^2 x = (\sqrt{3})^2 = 3 \]
Now we can find \( \cos^2 x + \cot^2 x \):
\[ \cos^2 x + \cot^2 x = \frac{3}{4} + 3 = \frac{3}{4} + \frac{12}{4} = \frac{15}{4} \]

### Step 5: Calculate \( \cos 3x \)
Using the angle \( x = 30^\circ \):
\[ \cos 3x = \cos(3 \times 30^\circ) = \cos 90^\circ = 0 \]

### Step 6: Calculate \( \sin 2x \)
Using the angle \( x = 30^\circ \):
\[ \sin 2x = \sin(2 \times 30^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2} \]

### Final Answers:
(i) \( x = 30^\circ \)
(ii) \( \cos^2 x + \cot^2 x = \frac{15}{4} \)
(iii) \( \cos 3x = 0 \)
(iv) \( \sin 2x = \frac{\sqrt{3}}{2} \)