Third Solution: Find $k$ if $k\sin 6x=\sin 2x$ given $\dfrac{\cos 6x}{\cos 2x}=\dfrac{1}{6}$.

Find $k$ if $k\sin 6x=\sin 2x$ given $\dfrac{\cos 6x}{\cos 2x}=\dfrac{1}{6}$.

Third method:

Third method revolves around the concept of Componendo and Dividendo, that says if $\dfrac{a}{b}=\dfrac{c}{d}$, then $\dfrac{a\pm b}{b}=\dfrac{c\pm d}{d}$.

Besides, we have to be super familiar with the triple angle formula for both sine and cosine function:

$\displaystyle \color{yellow}\bbox[5px,purple]{\sin 3x=3\sin x-4\sin^3 x}$ and $\displaystyle \color{black}\bbox[5px,orange]{\cos 3x=4\cos^3 x-3\cos x}$

Also, the s

Let's get started:

$\dfrac{\cos 6x}{\cos 2x}=\dfrac{1}{6}$

$\dfrac{\cos 6x-\cos 2x}{\cos 2x}=\dfrac{1-6}{6}$

$\dfrac{-2\sin \left(\dfrac{6x+2x}{2}\right)\sin \left(\dfrac{6x-2x}{2}\right)}{\cos 2x}=\dfrac{-5}{6}$

$\dfrac{2\sin 4x\sin 2x}{\cos 2x}=\dfrac{5}{6}$

$\dfrac{2(2\sin 2x \cos 2x) \sin 2x}{\cos 2x}=\dfrac{5}{6}$

$\dfrac{4(\sin 2x \cancel{\cos 2x})}{\cancel{\cos 2x}}\cdot \dfrac{\sin 2x}{\sin 2x}=\dfrac{5}{6}$

$\dfrac{4\sin^3 2x}{\sin 2x}=\dfrac{5}{6}$

$\dfrac{-4\sin^3 2x}{\sin 2x}=\dfrac{-5}{6}$

$\dfrac{-4\sin^3 2x+\sin 2x}{\sin 2x}=\dfrac{-5+6}{6}$

$\dfrac{-4\sin^3 2x+\sin 2x+\sin 2x}{\sin 2x}=\dfrac{-5+6+6}{6}$

$\dfrac{-4\sin^3 2x+\sin 2x+\sin 2x+\sin 2x}{\sin 2x}=\dfrac{5+6+6+6}{6}$

$\dfrac{3\sin 2x-4\sin^3 2x}{\sin 2x}=\dfrac{13}{6}$

$\dfrac{\sin 6x}{\sin 2x}=\dfrac{13}{6}$

and we're done.