Question

In a triangle ABC angle A is greater than angle B . if the measure 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

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

Answer 1

The correct option is C $\frac{2\pi }{3}$

Using the identity that: $\sin \left(3\theta \right)=3\sin \theta -4{\sin }^3\theta$
The equation can be written as:
$\sin \left(3x\right)=k$
As $A$ and $B$ are both solutions:
$\sin \left(3A\right)=\sin \left(3B\right)$
As $A\ne B$, the other solution is given by:
$3A=\pi -3B⟹A+B=\frac{\pi }{3}$
As $A,B,C$ are angles of the same triangle:
$A+B+C=\pi ⟹C=\pi -(A+B)=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$