Question

In a triangle $ABC$, 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)

$3\sin x-4{\sin }^{3}x=k0<k<1$
$\sin 3x=k$
Also $\sin 3A=k$
$\sin 3B=k$
$\Rightarrow 0<3A<\pi 0<3B<\pi$ ......(i)
Also $\sin 3A-\sin 3B=0$
$\Rightarrow 2\cos \frac{3}{2}(A+B)\sin \frac{3}{2}(A-B)=0$
$\cos \frac{3}{2}(A+B)=0\sin \frac{3}{2}(A-B)=0$....(ii)
Given $A>B$......(iii)
$\Rightarrow \sin \frac{3}{2}(A-B)\ne 0$
$\Rightarrow \cos \left(3\frac{(A+B)}{2}\right)=0$
$A+B=\frac{\pi }{3}C=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$