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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Solve the equation $\sin mx+\sin nx=0$

$\begin{array}{1 1}(A)\;\large\frac{2p\pi}{m+n},\frac{(2q+1)\pi}{m-n}&(B)\;\large\frac{6p\pi}{m+n},\frac{(3q+1)\pi}{2}\\(C)\;\large\frac{p\pi}{m+n},\frac{(7q+1)\pi}{4}&(D)\;\large\frac{8p\pi}{m+n},\frac{(2q+5)\pi}{2}\end{array} $

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1 Answer

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Toolbox:
  • $\sin \theta=0\Rightarrow \theta=n\pi$
  • $\sin A+\sin B=\sin \large\frac{A+B}{2}.$$\cos \large\frac{A+B}{2}$
$\sin mx+\sin nx=0$
$\Rightarrow 2\sin \large\frac{(m+n)x}{2}.$$\cos \large\frac{(m-n)x}{2}$$=0$
$\Rightarrow \sin\large\frac{(m+n)x}{2}$$=0$ or $\cos \large\frac{(m-n)x}{2}$$=0$
$\Rightarrow (\large\frac{m+n}{2})$$x=p\pi$ or $(\large\frac{m-n}{2})$$x=(2q+1)\large\frac{\pi}{2}$$\quad p,q\in I$
$\Rightarrow x=\large\frac{2p\pi}{m+n}$ or $x=\large\frac{(2q+1)\pi}{m-n}$$\quad p,q\in I$
Hence (A) is the correct answer.
answered May 22, 2014 by sreemathi.v
 
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