In a $Δ A B C, ∠ A > ∠ B .$ If $sin A$ and $sin B$ satisfy the equation $3 sin x - 4 {sin}^{3} x - k = 0, 0 < k < 1,$ then $∠ C$ is

\frac{π}{3}

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\frac{π}{2}

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\frac{2 π}{3}

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\frac{5 π}{6}

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Solution

The correct option is B $\frac{2 π}{3}$
$3 sin x - 4 {sin}^{3} x - k = 0 \Rightarrow sin 3 x = k$
Given $sin A, sin B$ satisfy this equation,
$sin 3 A = sin 3 B \Rightarrow 3 A = 3 B$ or $π - 3 B$
$\Rightarrow A = B$ or $A + B = \frac{π}{3} \Rightarrow ∠ C = \frac{2 π}{3}$

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$3 sin x - 4 {sin}^{3} x - k = 0 \Rightarrow sin 3 x = k$
Given $sin A, sin B$ satisfy this equation,
$sin 3 A = sin 3 B \Rightarrow 3 A = 3 B$ or $π - 3 B$
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