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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Find the general solution for each of the following equation $\cos 3x+\cos x-\cos 2x=0$

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

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1 Answer

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  • $\cos x+\cos y=2\cos \large\frac{x+y}{2}$$\cos \large\frac{x-y}{2}$
  • $\cos x=0\therefore x=(2n+1)\large\frac{\pi}{2}$
  • $\cos x=\cos y$
  • $x=2n\pi\pm y$
$\cos 3x+\cos x-\cos 2x=0$
$2\cos \large\frac{3x+x}{2}$$\cos \large\frac{3x-x}{2}$$-\cos 2x=0$
$2\cos 2x.\cos x-\cos 2x=0$
$\cos 2x(2\cos x-1)=0$
$\therefore$ If $\cos 2x=0,2x=(2n+1)\large\frac{\pi}{2}$
$\Rightarrow x=(2n+1)\large\frac{\pi}{4}$
$\therefore$ If $2\cos x-1=0,\cos x=\large\frac{1}{2}$$=\cos 60^{\large\circ}=\cos \large\frac{\pi}{3}$
$\therefore x=2n\pi\pm \large\frac{\pi}{3}$
Hence (A) is the correct answer.
answered Apr 18, 2014 by sreemathi.v
 

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