If `cos3A=(1)/(2)`, then how many value can sin A assume ?
`(0ltAlt360^(@))`

Options

A. 3
B. 4
C. 5
D. 6

Explanation

To solve the problem, we need to find how many values \( \sin A \) can assume given that \( \cos 3A = \frac{1}{2} \) and \( 0 < A < 360^\circ \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \cos 3A = \frac{1}{2} \]

2. **Identify the angles for which cosine equals \( \frac{1}{2} \):**
The cosine function equals \( \frac{1}{2} \) at:
\[ 3A = 60^\circ + 360^\circ n \quad \text{and} \quad 3A = 300^\circ + 360^\circ n \]
where \( n \) is any integer.

3. **Solve for \( A \) from the first equation:**
\[ 3A = 60^\circ + 360^\circ n \implies A = \frac{60^\circ + 360^\circ n}{3} = 20^\circ + 120^\circ n \]

4. **Solve for \( A \) from the second equation:**
\[ 3A = 300^\circ + 360^\circ n \implies A = \frac{300^\circ + 360^\circ n}{3} = 100^\circ + 120^\circ n \]

5. **Determine the values of \( A \) within the range \( 0 < A < 360^\circ \):**

- From \( A = 20^\circ + 120^\circ n \):
- For \( n = 0 \): \( A = 20^\circ \)
- For \( n = 1 \): \( A = 140^\circ \)
- For \( n = 2 \): \( A = 260^\circ \)
- For \( n = 3 \): \( A = 380^\circ \) (not valid)

- From \( A = 100^\circ + 120^\circ n \):
- For \( n = 0 \): \( A = 100^\circ \)
- For \( n = 1 \): \( A = 220^\circ \)
- For \( n = 2 \): \( A = 340^\circ \)
- For \( n = 3 \): \( A = 460^\circ \) (not valid)

6. **List all valid values of \( A \):**
- From the first equation: \( 20^\circ, 140^\circ, 260^\circ \)
- From the second equation: \( 100^\circ, 220^\circ, 340^\circ \)

Thus, the valid values of \( A \) are:
\[ 20^\circ, 100^\circ, 140^\circ, 220^\circ, 260^\circ, 340^\circ \]

7. **Count the number of values:**
There are a total of 6 valid values for \( A \).

### Conclusion:
Since \( \sin A \) will take the same values for each corresponding \( A \), we conclude that \( \sin A \) can assume **6 values**.

### Final Answer:
Thus, the number of values that \( \sin A \) can assume is **6**.

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