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Total number of solutions of $\cos x.\cos 2x.\cos 3x=\large\frac{1}{4}$ in $[0,\pi]$ is equal to


1 Answer

$\cos x.\cos 2x.\cos 3x=\large\frac{1}{4}$
$4\cos 3x.\cos x.\cos 2x=1$
$2(\cos 4x+\cos 2x)\cos 2x=1$
$2(2\cos^22x-1+\cos 2x)\cos 2x=1$
$4\cos^32x+2\cos^22x-2\cos 2x-1=0$
$2\cos^22x(2\cos 2x+1)-(2\cos 2x+1)=0$
$(2\cos 2x+1)(2\cos^22x-1)=0$
$\cos 4x(2\cos 2x+1)=0$
$\cos 4x=0$
$\cos 2x=-\large\frac{1}{2}$
$2x=2n_2\pi\pm \large\frac{2\pi}{3}$
Thus there are 6 solutions.
Hence (b) is the correct answer.
answered Nov 19, 2013 by sreemathi.v

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