If $2 \cos x + 2 \cos 3 x = \cos y$ and $2 \sin x + 2 \sin 3 x = \sin y$ then $\cos 2 x =$

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- \frac{1}{8}

\frac{7}{8}

\frac{1}{8}

- \frac{7}{8}

answer is A.

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Detailed Solution

2 \cos x + 2 \cos 3 x = \cos y

$\Rightarrow 2 [\cos x + \cos 3 x] = \cos y$

$\Rightarrow 2.2 \cos 2 x . \cos x = \cos y$

$\Rightarrow 4 \cos 2 x . \cos x = \cos y \to (1)$

$\Rightarrow 2 \sin x + 2 \sin 3 x = \sin y$

$\Rightarrow 2 (\sin x + \sin 3 x) = \sin y$

$\Rightarrow 4 \sin 2 x . \cos x = \sin y \to (2)$

$(1) \div (2)$

$\cot 2 x = \cot y \Rightarrow 2 x = y \Rightarrow \cos 2 x = \cos y$

$\Rightarrow \cos 2 x = \cos y = 2 \cos x + 2 \cos 3 x$ $= 4 \cos 2 x . \cos x$

$\Rightarrow \cos 2 x (1 - 4 \cos x) = 0$

$\Rightarrow 1 - 4 \cos x = 0 \Rightarrow \cos x = \frac{1}{4}$

$\Rightarrow \cos 2 x = 2 \cos^{2} x - 1$

$= 2 (\frac{1}{16}) - 1$

$= \frac{1}{8} - 1 = - \frac{7}{8}$

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