Question

In a triangle ABC, the angle A is greater than the angle B. If the values of the angles A and B satisfy the equation 3sinx - 4sin³x - k =0, 0 <k< 1, then the measure of angle C is

• A. $\frac{\pi}{3}$
• B. $\frac{\pi}{2}$
• C. $\frac{2\pi}{3}$
• D. $\frac{5\pi}{6}$

Answer 1

Answer: (C)

$\frac{2\pi}{3}$

Answer 2

The given equation can be written as $\sin 3x = k,\text{ }0 < k < 1$

Since A and B satisfy this equation

$0 < 3A,$ 3B < π as 0 < k < 1

Also $\sin 3A = k = \sin 3B$ ⇒ $\sin 3A - \sin 3B = 0$

⇒ $2\cos\frac{3(A + B)}{2}\sin\frac{3(A - B)}{2} = 0$⇒ either $\cos\frac{3(A + B)}{2} = 0$ or $\sin\frac{3(A - B)}{2} = 0$

But $\sin\frac{3(A - B)}{2} \neq 0$ as A > B and 0 < 3A, 3B < π so $\cos\frac{3(A + B)}{2} = 0$

⇒ $\cos\frac{3}{2}(\pi - C) = 0$

⇒ $\sin\frac{3C}{2} = 0$ ⇒ $C = \frac{2\pi}{3}$.