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Find the general solution for each of the following equation $\sin x+\sin 3x+\sin 5x=0$

$\begin{array}{1 1}(A)\;n\pi\pm \large\frac{\pi}{6}&(B)\;n\pi\pm \large\frac{\pi}{3}\\(C)\;2n\pi\pm \large\frac{\pi}{2}&(D)\;n\pi\pm \large\frac{\pi}{4}\end{array} $

1 Answer

  • $\sin A+\sin B=2\sin \large\frac{A+B}{2}$$\cos \large\frac{A-B}{2}$
$\sin x+\sin 3x+\sin 5x=0$
$(\sin 5x+\sin x)+\sin 3x=0$
$2\sin \large\frac{5x+x}{2}$$\cos \large\frac{5x-x}{2}$$+\sin 3x=0$
$2\sin 3x\cos 2x+\sin 3x=0$
$\Rightarrow \sin 3x(2\cos 2x+1)=0$
If $\sin 3x=0,3x=n\pi$ or $x=\large\frac{n\pi}{3}$
If $2\cos 2x+1=0,\cos 2x=-\large\frac{1}{2}$$=\cos (\pi-\large\frac{\pi}{3})$
$\Rightarrow \cos \large\frac{2\pi}{3}$
$2x=2n\pi+\large\frac{2\pi}{3}$ or $x=n\pi\pm \large\frac{\pi}{3}$
Hence (B) is the correct answer.
answered Apr 18, 2014 by sreemathi.v

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