First slide

Multiple and sub- multiple Angles

Question

In a triangle ABC, angle A is greater than B, if the measures 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

Moderate

A

$\frac{- π}{2}$
B

$\frac{π}{3}$
C

$\frac{2 π}{3}$
D

$\frac{4 π}{3}$

Solution

$\begin{array}{l} G i v e n A > B a n d, A a n d B s a t i s f y t h e e q u a t i o n \\ 3 \sin x - 4 \sin^{3} x - k = 0 \\ \Rightarrow \sin 3 x - k = 0 \\ (o r) \sin 3 A - k = 0 a n d \sin 3 B - K = 0 \\ ∴ \sin 3 A = \sin 3 B = \sin (π - 3 B) \\ ⇒ 3 A = π - 3 B (∵ A > B) \\ ⇒ A + B = \frac{π}{3} T h e n C = π - \frac{π}{3} = \frac{2 π}{3} \end{array}$

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