In a triangle ABC ,angle A is greater than B.If the measures of angles Aand B satisfy the equation `3sinx-4 sin^(3)x-k=0,0ltklt1`, then the measure of angle C, is

Options

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

Answer

To solve the problem step by step, we will analyze the given information and derive the required angle \( C \) in triangle \( ABC \).

### Step 1: Understand the given equation
The equation provided is:
\[ 3 \sin x - 4 \sin^3 x - k = 0 \]
This can be rearranged to:
\[ 3 \sin x - 4 \sin^3 x = k \]

### Step 2: Recognize the trigonometric identity
Notice that the left-hand side can be expressed using the triple angle formula for sine:
\[ 3 \sin x - 4 \sin^3 x = \sin 3x \]
Thus, we can rewrite the equation as:
\[ \sin 3x = k \]

### Step 3: Relate angles A and B
Given that \( A > B \), we can set:
\[ k = \sin 3A \quad \text{and} \quad k = \sin 3B \]
This implies:
\[ \sin 3A = \sin 3B \]

### Step 4: Use the sine property
Since \( \sin 3A = \sin 3B \), we have two cases:
1. \( 3A = 3B + n \cdot 180^\circ \) for some integer \( n \)
2. \( 3A = 180^\circ - 3B + n \cdot 180^\circ \)

Given that \( A > B \), we will use the second case:
\[ 3A = 180^\circ - 3B \]
Rearranging gives:
\[ 3A + 3B = 180^\circ \]
Dividing through by 3:
\[ A + B = 60^\circ \]

### Step 5: Find angle C
In any triangle, the sum of the angles is \( 180^\circ \):
\[ A + B + C = 180^\circ \]
Substituting \( A + B = 60^\circ \):
\[ 60^\circ + C = 180^\circ \]
Thus, solving for \( C \):
\[ C = 180^\circ - 60^\circ = 120^\circ \]

### Final Answer
The measure of angle \( C \) is:
\[ \boxed{120^\circ} \]