Solve the following equations for A, if :
(i) `sin 3 A = (sqrt(3))/(2)`
(ii) `sqrt(3) cot 2A = 1`

Answer

To solve the equations for \( A \), we will tackle each equation step by step.

### (i) Solve \( \sin 3A = \frac{\sqrt{3}}{2} \)

**Step 1:** Recognize the value of sine.
We know that \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).

**Hint:** Recall the unit circle values for sine at common angles.

**Step 2:** Set the angles equal.
Since \( \sin 3A = \sin 60^\circ \), we can set the angles equal to each other:
\[ 3A = 60^\circ + n \cdot 360^\circ \quad \text{or} \quad 3A = 180^\circ - 60^\circ + n \cdot 360^\circ \]
where \( n \) is any integer.

**Hint:** Use the property of sine that states \( \sin x = \sin(180^\circ - x) \).

**Step 3:** Solve for \( A \).
From \( 3A = 60^\circ \):
\[ A = \frac{60^\circ}{3} = 20^\circ \]

From \( 3A = 120^\circ \):
\[ A = \frac{120^\circ}{3} = 40^\circ \]

**Hint:** Divide by 3 to isolate \( A \).

**Step 4:** General solution.
The general solutions for \( A \) can be expressed as:
\[ A = 20^\circ + n \cdot 120^\circ \quad \text{and} \quad A = 40^\circ + n \cdot 120^\circ \]
for any integer \( n \).

### (ii) Solve \( \sqrt{3} \cot 2A = 1 \)

**Step 1:** Rearrange the equation.
We can rewrite the equation as:
\[ \cot 2A = \frac{1}{\sqrt{3}} \]

**Hint:** Remember that \( \cot \theta = \frac{1}{\tan \theta} \).

**Step 2:** Identify the angle.
We know that \( \cot 60^\circ = \frac{1}{\sqrt{3}} \).

**Hint:** Recall the values of cotangent at common angles.

**Step 3:** Set the angles equal.
Thus, we have:
\[ 2A = 60^\circ + n \cdot 360^\circ \quad \text{or} \quad 2A = 180^\circ - 60^\circ + n \cdot 360^\circ \]

**Step 4:** Solve for \( A \).
From \( 2A = 60^\circ \):
\[ A = \frac{60^\circ}{2} = 30^\circ \]

From \( 2A = 120^\circ \):
\[ A = \frac{120^\circ}{2} = 60^\circ \]

**Hint:** Divide by 2 to isolate \( A \).

**Step 5:** General solution.
The general solutions for \( A \) can be expressed as:
\[ A = 30^\circ + n \cdot 180^\circ \quad \text{and} \quad A = 60^\circ + n \cdot 180^\circ \]
for any integer \( n \).

### Summary of Solutions
From the first equation, we found:
- \( A = 20^\circ \) or \( A = 40^\circ \)

From the second equation, we found:
- \( A = 30^\circ \) or \( A = 60^\circ \)