Question

In a triangle ABC, angle A is greater than B. If the measures of angles A and B satisfy the equation $3sinx–4si{n}^{3}x=k,0<k<1$, then the value 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

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

$3sinx-4si{n}^{3}x=ksin3x=k\Rightarrow sin3A=k,sin3B=k\Rightarrow sin3A-sin3B=0\Rightarrow 2sin\left(\frac{3A-3B}{2}\right)cos\left(\frac{3A+3B}{2}\right)=0\Rightarrow cos\frac{3}{2}(A+B)=0\Rightarrow A+B=\frac{\pi }{3}\Rightarrow C=\frac{2\pi }{3}$