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In a triangle ABC, the angle A is greater than angle B. If the measures of angles A and B satisfy the equation $3 \sin x - 4 \sin^{3} x - k = 0, 0 < k < 1$ , then the angle C is

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$\frac{π}{3}$

$\frac{π}{4}$

$\frac{2 π}{3}$

$\frac{π}{2}$

answer is B.

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Detailed Solution

$3 \sin x - 4 \sin^{3} x = k$

sin 3x = k

$∴$ sin 3A = sin 3B or sin (180° - 3B),

since A> B

3A = 180° - 3B

A= 60° - B

$\begin{array}{l} ∴ A + B = 60^{\circ} \\ C = 120^{\circ} = \frac{2 π}{3} . \end{array}$

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