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
π4
B
π2
C
2π3
D
5π6

Solution

## The correct option is C $$\dfrac{2\pi}3$$$$3\sin {x} - 4 \sin^{3} x = k \Rightarrow \sin {3x} = k$$As $$A$$ and $$B$$ satisfy this, we get$$\sin {3A} = k$$ and $$\sin {3B} = k$$$$\Rightarrow \sin {3A} = \sin {3B}$$$$\Rightarrow \sin {3A} - \sin {3B} = 0$$$$\Rightarrow 2 \cos \left (\dfrac {3A + 3B}{2}\right ) \sin \left (\dfrac {3A - 3B}{2}\right ) = 0$$Using $$A + B + C = \pi \Rightarrow A + B = \pi - C$$$$\cos \left (3 \left (\dfrac {\pi - C}{2}\right )\right ) \sin \left (\dfrac {3A - 3B}{2}\right ) = 0$$Now as $$A > B \Rightarrow \sin \left (\dfrac {3A - 3B}{2} \right ) > 0$$$$\sin \left (\dfrac {3C}{2}\right ) = 0 \Rightarrow \dfrac {3C}{2} = n\pi \Rightarrow C = \dfrac {2 n \pi}{3}$$For $$n = 1$$$$C = \dfrac {2\pi}{3}$$Mathematics

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