In triangle ABC, the angle B is greater than angle C. If the measure of angles B and C satisfy the equation $4 \sin^{3} x - 3 \sin x + 0.75 = 0$ , then measue of angles A is

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$\frac{π}{6}$

$\frac{5 π}{6}$

$\frac{π}{2}$

$\frac{2 π}{3}$

answer is B.

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Detailed Solution

$\begin{array}{l} 4 \sin^{3} x - 3 \sin x + 0.75 = 0 \\ 0.75 = \sin 3 x \\ \sin 3 x = \frac{3}{4} \to (1) \\ \sin 3 B = \frac{3}{4}; \sin 3 C = \frac{3}{4} \end{array}$

$\begin{array}{l} \sin 3 B = \sin 3 C \\ \Rightarrow B = C or 180 - 3 C = 3 B \end{array}$

But $B > C \Rightarrow B \neq C$

$3 B = 180 - 3 C$

$\begin{array}{l} 3 (B + C) = 180 \\ B + C = 60 \\ 180 - A = 60 \\ A = 120 = \frac{2 π}{3} \end{array}$

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