Find $k$ if $k\sin 6x=\sin 2x$ given $6\cos 6x=\cos 2x$.

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

There are at least three different methods of solving for this particular problem. One is rather straightforward, and the other two methods are quite special.

We will discuss each of them separately, let's first talk about the straightforward method:

Note that we have, from the triple angle formula for sine and cosine function that $\displaystyle \color{yellow}\bbox[5px,purple]{\sin 3x=3\sin x-4\sin^3 x}$ and $\displaystyle \color{yellow}\bbox[5px,green]{\cos 3x=4\cos^3 x-3\cos x}$:

$k\sin 6x=\sin 2x$

$\begin{align*}k&=\dfrac{\sin 2x}{\sin 6x}\\&=\dfrac{\sin 2x}{3\sin 2x-4\sin^3 2x}\\&=\dfrac{\sin 2x}{\sin 2x(3-4\sin^2 2x)}\\&=\dfrac{1}{3-4(1-\cos^2 2x)}\\&=\dfrac{1}{-1+4\cos^2 2x}\\&=\dfrac{1}{4\cos^2 2x-1}\end{align*}$

Thus, if we have found the value for $\cos 2x$, then our mission is accomplished.

$6\cos 6x=\cos 2x$

$6(4\cos^3 2x-3\cos 2x)=\cos 2x$

$6\cos 2x(4\cos^2 2x-3)=\cos 2x$

$\cos 2x(6(4\cos^2 2x-3)-1)=0$

$\cos 2x(24\cos^2 2x-19)=0$

Therefore, we have either

$\cos 2x=0$ or

$24\cos^2 2x-19=0$ which implies $\cos 2x=\pm\sqrt{\dfrac{19}{24}}$.

Therefore, taking $\cos 2x=0$ we get:

$\begin{align*}k&=\dfrac{1}{4\cos^2 2x-1}\\&=-1\end{align*}$

Hey, we can't take $\cos 2x=0$ because $\cos 2x$ is in the denominator, and a zero denominator will make the entire equation undefined.

So, when taking $\cos 2x=\pm\sqrt{\dfrac{19}{24}}$, it gives us the only solution for this problem, which is:

$\begin{align*}k&=\dfrac{1}{4\left(\dfrac{19}{24}\right)-1}\\&=\dfrac{6}{13}\end{align*}$