Question

In a triangle $A B C$, angle $A$ is greater than angle $B$. If the measures of angles $A$ and $B$ satisfy the equation $3 \sin x-4 \sin ^3 x-k=0,0< k< 1$, then the measure of angle $C$ is :

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

Answer 1

Answer: (C)

$3 \sin x-4 \sin ^3 x=k 0 < k< 1$
$\sin 3 x=k$
Also $ \sin 3 A=k$
$\sin 3 B=k$
$\Rightarrow 0< 3 A< \pi 0< 3 B< \pi$ ......(i)
Also $ \sin 3 A-\sin 3 B=0$
$\Rightarrow 2 \cos \frac{3}{2}(A+B) \sin \frac{3}{2}(A-B)=0$
$\cos \frac{3}{2}(A+B)=0 \sin \frac{3}{2}(A-B)=0$....(ii)
Given $A >B$......(iii)
$\Rightarrow \sin \frac{3}{2}(A-B) \neq 0 $
$ \Rightarrow \cos \left(3 \frac{(A+B)}{2}\right)=0$
$A+B=\frac{\pi}{3} C=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}$