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 :
$$\frac{2\pi }{3}$$
Solution :
$$\eqalign{
& {\text{Given that}}\,{\text{ }}A > B \cr
& {\text{and }}3\sin x - 4{\sin ^3}x - k = 0,\,\,\,\,\,\,0 < k < 1 \cr
& \Rightarrow \,\,\sin \,3x = k \cr} $$
As $$A$$ and $$B$$ satisfy above eq. (given)
$$\eqalign{
& \therefore \,\,\sin \,3A\, = k,\,\,\sin \,3B = k \cr
& \Rightarrow \,\sin \,3A - \sin \,3B = 0 \cr
& \Rightarrow \,\,2\,\cos \,\frac{{3A + 3B}}{2}\sin \,\frac{{3A - 3B}}{2} = 0 \cr
& \Rightarrow \,\,\cos \left( {\frac{{3A + 3B}}{2}} \right) = 0\,\,{\text{or }}\sin \left( {\frac{{3A - 3B}}{2}} \right) = 0 \cr
& \Rightarrow \,\,\frac{{3A + 3B}}{2} = {90^ \circ }\,\,{\text{or }}\frac{{3A - 3B}}{2} = 0 \cr
& \Rightarrow \,\,A + B = {60^ \circ }\,\,{\text{or }}A = B \cr} $$
But given that $$A > B,\,\,\therefore \,\,A \ne B$$
Thus, $$A + B = 60°$$
But $$A + B + C = 180°$$
⇒ $$C = 180° - 60° = 120°$$
∴ $$C = \frac{{2\pi }}{3}$$