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UPSEE 2018 Solved Paper

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Question : 106 of 150

Marks: +1, -0

If

\cos 3x \cos 2x \cos x = \frac{1}{4}

0 < x < \frac{\pi}{4}

, then the value of

x

is

$\frac{\pi}{6}$
$\frac{\pi}{5}$
$\frac{\pi}{8}$
$\frac{\pi}{7}$

Solution:

Given,

\cos 3x \cos 2x \cos x = \frac{1}{4}

\Rightarrow 2(2 \cos x \cos 3x) \cos 2x = 1

\Rightarrow 2(\cos 4x + \cos 2x) \cos 2x = 1

[\because 2 \cos A \cos B = \cos(A+B) + \cos(A-B)]

\Rightarrow 2(2 \cos^{2} 2x - 1 + \cos 2x) \cos 2x = 1

\Rightarrow 4 \cos^{3} 2x - 2 \cos 2x

+2 \cos^{2} 2x - 1 = 0

\Rightarrow 2 \cos 2x (2 \cos^{2} 2x - 1)

+1(2 \cos^{2} x - 1) = 0

\Rightarrow (2 \cos^{2} 2x - 1)(2 \cos 2x + 1) = 0

\Rightarrow \cos 4x (2 \cos 2x + 1) = 0

\Rightarrow \cos 4x = 0

or

2 \cos 2x + 1 = 0

\Rightarrow \cos 4x = \cos \frac{\pi}{2}

or

\cos 2x = \frac{-1}{2}

\Rightarrow 4x = \frac{\pi}{2}

or

\cos 2x = \cos \frac{2\pi}{3}

⇒

x = \frac{\pi}{8}

or

2x = \frac{2\pi}{3}

⇒

x = \frac{\pi}{8}

or

x = \frac{\pi}{3}

Since,

x \in \left(0, \frac{\pi}{4}\right)

\therefore x = \frac{\pi}{8}

is the required value.

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