In the `Delta ABCA gt B.` If the measures of A and B satisfy the equation `3 sin x - 4 sin ^(3) x - k =0, 0 lt k lt 1.` Then the measure of C is

Options

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

Answer

To solve the problem, we need to find the measure of angle C in triangle ABC given that the measures of angles A and B satisfy the equation \(3 \sin x - 4 \sin^3 x - k = 0\) where \(0 < k < 1\).

### Step-by-Step Solution:

1. **Rearranging the Equation**:
We start with the equation:
\[ 3 \sin x - 4 \sin^3 x = k \]
This can be recognized as the formula for \(\sin 3x\):
\[ \sin 3x = 3 \sin x - 4 \sin^3 x \]
Therefore, we can rewrite the equation as:
\[ \sin 3x = k \]

2. **Understanding the Relationship**:
Since we have \(A\) and \(B\) such that:
\[ \sin 3A = k \quad \text{and} \quad \sin 3B = k \]
This implies that:
\[ 3A = n\pi + (-1)^n \arcsin(k) \quad \text{for some integer } n \]
and
\[ 3B = m\pi + (-1)^m \arcsin(k) \quad \text{for some integer } m \]

3. **Finding Possible Cases**:
The sine function has the property that:
\[ \sin \theta = \sin(\pi - \theta) \]
Thus, we have two cases:
- Case 1: \(3A = 3B\) which implies \(A = B\) (not possible since \(A > B\))
- Case 2: \(3A = \pi - 3B\)

4. **Solving for C**:
From Case 2, we have:
\[ 3A + 3B = \pi \]
Dividing through by 3 gives:
\[ A + B = \frac{\pi}{3} \]
Since the sum of angles in a triangle is \(180^\circ\) or \(\pi\) radians, we can find angle C:
\[ C = \pi - (A + B) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]

5. **Conclusion**:
Therefore, the measure of angle C is:
\[ C = \frac{2\pi}{3} \]