In triangle `ABC`, angle `A` is greater than angle `B`. If the measure of angles `A` and `B` satisfy the equation `3sinx-4sin^3x-k=0`. Find the value of angle `C`.

Options

A. `pi/3`
B. `pi/2`
C. `(2pi)/3`
D. `(5pi)/6`

Explanation

To solve the problem, we need to find the value of angle \( C \) in triangle \( ABC \) given that angle \( A \) is greater than angle \( B \) and that the angles satisfy the equation \( 3\sin x - 4\sin^3 x - k = 0 \).

### Step-by-Step Solution:

1. **Understanding the Equation**:
The equation given is \( 3\sin x - 4\sin^3 x - k = 0 \). This can be rearranged to:
\[ 3\sin x - 4\sin^3 x = k \]

2. **Using the Identity**:
We can use the identity for \( \sin 3x \):
\[ \sin 3x = 3\sin x - 4\sin^3 x \]
Therefore, we can rewrite the equation as:
\[ \sin 3x = k \]

3. **Finding the Angles**:
Since \( k \) is a value between 0 and 1 (as \( \sin \) values range between -1 and 1), we can express:
\[ 3x = \arcsin(k) \]
This implies that:
\[ x = \frac{1}{3} \arcsin(k) \]

4. **Relating Angles A and B**:
Since we have \( A \) and \( B \) as angles in triangle \( ABC \) with \( A > B \), we can express:
\[ \sin 3A = k \quad \text{and} \quad \sin 3B = k \]
This leads to:
\[ 3A = \arcsin(k) \quad \text{and} \quad 3B = \arcsin(k) \]
Since \( A > B \), we can conclude:
\[ 3A > 3B \]

5. **Using the Sine Difference**:
From the sine difference identity, we have:
\[ \sin 3A - \sin 3B = 0 \]
This implies:
\[ 3A - 3B = n\pi \quad \text{for some integer } n \]

6. **Finding C**:
In a triangle, the sum of the angles is \( 180^\circ \):
\[ A + B + C = 180^\circ \]
Since \( A + B = 180^\circ - C \), we can substitute:
\[ C = 180^\circ - (A + B) \]

7. **Conclusion**:
Given that \( A + B = 90^\circ \) (as derived from \( 3A + 3B = 90^\circ \)), we find:
\[ C = 180^\circ - 90^\circ = 90^\circ \]

Thus, the value of angle \( C \) is \( 90^\circ \).