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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Find the general solution for each of the following equation $\cos 4x=\cos 2x$

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

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1 Answer

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  • $\cos x-\cos y=-2\sin\large\frac{x+y}{2}$$\sin \large\frac{x-y}{2}$
$\cos 4x=\cos 2x$
$\cos 2x-\cos 4x=0$
$-2\sin \large\frac{2x+4x}{2}$$.\sin \large\frac{2x-4x}{2}$$=0$
$-2\sin 3x.\sin(-x)=0$
$2\sin 3x.\sin x=0$
If $\sin 3x=0$
$\Rightarrow 3x=n\pi$
$\Rightarrow x=\large\frac{n\pi}{3}$
$\sin x=0$
$\Rightarrow x=n\pi$ where $n\in z$
Hence (A) is the correct answer.
answered Apr 18, 2014 by sreemathi.v

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