Question

If $3\cos x+2\cos 3x=\cos y,3\sin x+2\sin 3x=\sin y,$ then the value of $\cos 2x$ is

• A. $-1$
• B. $\frac{1}{8}$
• C. $-\frac{1}{8}$
• D. $\frac{7}{8}$

Answer 1

The correct option is A $-1$

Given: $3\cos x+2\cos 3x=\cos y$

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

$\Rightarrow {\cos }^{2}y+{\sin }^{2}y={(3\cos x+2\cos 3x)}^2+{(3\sin x+2\sin 3x)}^2$

$\Rightarrow 1=9{\cos }^{2}x+4{\cos }^{2}3x+12\cos 3x\cos 5x+9{\sin }^{2}x+4{\sin }^{2}3x+12\sin x\sin 3x$

............. $[sin²\theta +cos²\theta =1]$

$\Rightarrow 1=9({\cos }^{2}x+{\sin }^{2}x)+12(\cos 3x\cos x+\sin x\sin 3x)+4({\cos }^{2}3x+{\sin }^{2}3x)$

$\Rightarrow 1=9+12\cos (3x-x)+4$ ........$\cos (a-b)=\cos a\cdot \cos b+\sin a\cdot sinb$

$\Rightarrow 1=13+12\cos 2x$

$\Rightarrow 12\cos 2x=-12$

$\Rightarrow \cos 2x=-1$

Hence, the answer is $-1.$