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

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$\frac{- π}{2}$

$\frac{4 π}{3}$

$\frac{π}{3}$

$\frac{2 π}{3}$

answer is C.

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Detailed Solution

$A > B,$ A and B are satisfy the equation
$\Rightarrow \sin 3 A - K = 0, \sin 3 B - K = 0$
$\sin 3 A = \sin 3 B$
$3 A = π - 3 B$
$A + B = \frac{π}{3}; C = \frac{2 π}{3}$

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