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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Find the general solution of the equation $\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x$

$\begin{array}{1 1}(A)\;x=\large\frac{n\pi}{2}+\frac{\pi}{6}&(B)\;x=\large\frac{n\pi}{2}+\frac{\pi}{8}\\(C)\;x=\large\frac{n\pi}{4}+\frac{\pi}{6}&(D)\;x=\large\frac{n\pi}{8}+\frac{\pi}{6}\end{array} $

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Toolbox:
  • $\sin A+\sin B=2\sin(\large\frac{A+B}{2})$$\cos (\large\frac{A-B}{2})$
  • $\cos A+\cos B=2\cos (\large\frac{A+B}{2})$$\cos (\large\frac{A-B}{2})$
$\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x$
$\sin 3x+\sin x-3\sin^2x =\cos 3x+\cos x-3\cos 2x$
$2\sin\large\frac{3x+x}{2}$$\cos\large\frac{3x-x}{2}$$-3\sin 2x=2\cos\large\frac{3x+x}{2}.$$\cos\large\frac{3x-x}{2}$$-3\cos 2x$
$2\sin 2x\cos x-3\sin 2x=2\cos 2x.\cos x-3\cos 2x$
$\sin 2x(2\cos x-3)=\cos 2x(2\cos x-3)$
$\large\frac{\sin 2x}{\cos 2x}$$=1$
$\tan 2x=1=\tan\large\frac{\pi}{4}$
$\therefore 2x=n\pi+\large\frac{\pi}{4}$
$x=\large\frac{n\pi}{2}+\frac{\pi}{8}$
Hence (B) is the correct answer.
answered Apr 21, 2014 by sreemathi.v
 

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