Trigonometrical Equations 1 Question 15
15. In a $\triangle A B C$, 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
(1990, 2M)
(a) $\frac{\pi}{3}$
(b) $\frac{\pi}{2}$
(c) $\frac{2 \pi}{3}$
(d) $\frac{5 \pi}{6}$
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Answer:
Correct Answer: 15. $(c)$
Solution:
- Given, $3 \sin x-4 \sin ^{3} x=k, 0<k<1$ which can also be written as $\sin 3 x=k$.
It is given that $A$ and $B$ are solutions of this equation. Therefore,
$ \begin{array}{lrl} & \sin 3 A=k \text { and } \sin 3 B=k, \text { where } 0<k<1 \\ \Rightarrow & 0<3 A<\pi \text { and } 0<3 B<\pi \\ \text { Now, } & \sin 3 A=k \text { and } \sin 3 B=k \\ \Rightarrow & \sin 3 A-\sin 3 B=0 \\ \Rightarrow & 2 \cos \frac{3}{2}(A+B) \sin \frac{3}{2}(A-B)=0 \end{array} $
$\Rightarrow \quad \cos 3 \frac{A+B}{2}=0, \sin 3 \frac{A-B}{2}=0$
But it is given that, $A>B$ and $0<3 A<\pi, 0<3 B<\pi$.
Therefore, $\sin 3 \frac{A-B}{2} \neq 0$
Hence, $\quad \cos 3 \frac{A+B}{2}=0$
$ \begin{array}{rlrl} \Rightarrow 3 \frac{A+B}{2} =\frac{\pi}{2} \\ \Rightarrow A+B =\frac{\pi}{3} \\ \Rightarrow C =\pi-(A+B)=\frac{2 \pi}{3} \end{array} $
