Question

The possible values of $\theta$ belongs to $\left(0,\mathrm{\pi }\right)$ such that $\sin \left(\theta \right)+\sin \left(4\theta \right)+\sin \left(7\theta \right)=0$ are

• A. $\raisebox{1ex}{$2\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$9$}\right.,\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.,\raisebox{1ex}{$4\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$9$}\right.,\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\raisebox{1ex}{$3\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.,\raisebox{1ex}{$8\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$9$}\right.$
• B. $\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.,\raisebox{1ex}{$5\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$12$}\right.,\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\raisebox{1ex}{$2\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.,\raisebox{1ex}{$3\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.,\raisebox{1ex}{$8\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$9$}\right.$
• C. $\raisebox{1ex}{$2\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$9$}\right.,\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.,\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\raisebox{1ex}{$2\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.,\raisebox{1ex}{$3\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.,\raisebox{1ex}{$35\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$36$}\right.$
• D. $\raisebox{1ex}{$2\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$9$}\right.,\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.,\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\raisebox{1ex}{$2\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.,\raisebox{1ex}{$3\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.,\raisebox{1ex}{$8\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$9$}\right.$

Answer 1

The correct option is A

$\raisebox{1ex}{$2\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$9$}\right.,\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.,\raisebox{1ex}{$4\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$9$}\right.,\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\raisebox{1ex}{$3\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.,\raisebox{1ex}{$8\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$9$}\right.$

Explanation for the correct option:

Step $1:$ Solve $\sin \left(\theta \right)+\sin \left(4\theta \right)+\sin \left(7\theta \right)=0$

$\sin \left(\theta \right)+\sin \left(4\theta \right)+\sin \left(7\theta \right)=0$

$\sin \left(4\theta -3\theta \right)+\sin \left(4\theta \right)+\sin \left(4\theta +3\theta \right)=0$

$[\sin \left(4\theta -3\theta \right)+\sin \left(4\theta +3\theta \right)]+\sin \left(4\theta \right)=0$

$2\sin 4\theta \cos 3\theta +\sin \left(4\theta \right)=0$ $\left[\because \sin (A-B)+\sin (A+B)=2\mathrm{sinAcosB}\right]$

$\sin 4\theta \left(2\cos 3\theta +1\right)=0$

$$\begin{gathered} \sin 4\theta =0,2\cos 3\theta +1=0 \\ \sin 4\theta =0,\cos 3\theta =-\frac{1}{2} \end{gathered}$$

Step $2:$ Solve for values of $\theta$.

As we know that when $\sin \theta =0\Rightarrow \theta =n\pi$ and when $\cos \theta =\cos \alpha \Rightarrow \theta =2n\pi \pm \alpha$

$$\begin{gathered} \sin 4\theta =0,\cos 3\theta =\cos \left(\frac{\mathrm{\pi }}{3}\right) \\ 4\theta =n\mathrm{\pi },3\mathrm{\theta }=(2\mathrm{n}+1)\mathrm{\pi }\pm \left(\frac{\mathrm{\pi }}{3}\right) \\ \mathrm{\theta }=\frac{n\mathrm{\pi }}{4},\mathrm{\theta }=\left[\frac{2n+1}{3}\right]\mathrm{\pi }\pm \frac{\mathrm{\pi }}{9} \end{gathered}$$

Thus,$0<\left(\frac{n\mathrm{\pi }}{4}\right)<\mathrm{\pi };\theta =\frac{2\mathrm{\pi }}{9},\frac{4\mathrm{\pi }}{9},\frac{8\mathrm{\pi }}{9}$

Therefore,$\theta =\frac{2\mathrm{\pi }}{9},\frac{\mathrm{\pi }}{4},\frac{4\mathrm{\pi }}{9},\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{4},\frac{8\mathrm{\pi }}{9}$

Hence the correct option is option(A)