$\sqrt{2} \cos \left(\dfrac{x}{5}-\dfrac{\pi }{12}\right)-\sqrt{6}\sin \left(\dfrac{x}{5}-\dfrac{\pi}{12}\right)=2\left(\sin \left((\dfrac{x}{5}-\dfrac{2\pi}{3}\right)-\sin \left(\dfrac{3x}{5}+\dfrac{\pi}{6}\right)\right)$
The LHS of the equation has been simplified down to
[MATH]\color{black}\bbox[5px,red]{\sqrt{2} \cos \left(\dfrac{x}{5}-\dfrac{\pi }{12}\right)-\sqrt{6}\sin \left(\dfrac{x}{5}-\dfrac{\pi}{12}\right)=2\cos \left(\dfrac{x}{5}\right)-2\sin \left(\dfrac{x}{5}\right)}[/MATH]
Whereas the RHS of the equation, after the failure to work on it using the angle sum and difference identities, we want to try the sum-to-product formula for sine function that says $\sin A-\sin B=2\cos \dfrac{A+B}{2}\sin \dfrac{A-B}{2}$.
[MATH]\color{black}\bbox[5px,orange]{2\left(\sin \left(\dfrac{x}{5}-\dfrac{2\pi}{3}\right)-\sin \left(\dfrac{3x}{5}+\dfrac{\pi}{6}\right)\right)}[/MATH]
$=2\left(2\cos \left(\dfrac{\dfrac{x}{5}-\dfrac{2\pi}{3}+\dfrac{3x}{5}+\dfrac{\pi}{6}}{2}\right)\sin \left(\dfrac{\dfrac{x}{5}-\dfrac{2\pi}{3}-\dfrac{3x}{5}-\dfrac{\pi}{6}}{2}\right)\right)$
$=2\left(2\cos \left(\dfrac{2x}{5}-\dfrac{\pi}{4}\right)\sin \left(-\dfrac{x}{5}-\dfrac{5\pi}{12}\right)\right)$
$=-4\left(\cos \left(\dfrac{2x}{5}-\dfrac{\pi}{4}\right)\sin \left(\dfrac{x}{5}+\dfrac{5\pi}{12}\right)\right)$
Putting these together, we see that:
$\color{black}\bbox[5px,red]{2\cos \left(\dfrac{x}{5}\right)-2\sin \left(\dfrac{x}{5}\right)}\color{black}=\color{black}\bbox[5px,orange]{-4\left(\cos \left(\dfrac{2x}{5}-\dfrac{\pi}{4}\right)\sin \left(\dfrac{x}{5}+\dfrac{5\pi}{12}\right)\right)}$
That doesn't seem like we could "merge" any of the term together, except for expanding the terms using the angle sum and difference formulas.
$2\cos \left(\dfrac{x}{5}\right)-2\sin \left(\dfrac{x}{5}\right)=-4\left(\cos \left(\dfrac{2x}{5}-\dfrac{\pi}{4}\right)\sin \left(\dfrac{x}{5}+\dfrac{5\pi}{12}\right)\right)$
$ 2\cos \left(\dfrac{x}{5}\right)-2\sin \left(\dfrac{x}{5}\right)$
$\small =-4\left(\cos \left(\dfrac{2x}{5}\right)\cos \left(\dfrac{\pi }{4}\right)+\sin \left(\dfrac{2x}{5}\right)\sin \left(\dfrac{\pi}{4}\right)\right)\left(\sin \left(\dfrac{x}{5}\right)\cos \left(\dfrac{5\pi }{12}\right)+\cos \left(\dfrac{x}{5}\right)\sin \left(\dfrac{5\pi}{12}\right)\right)$
$\small =-4\left(\dfrac{\sqrt{2}}{2}\right)\left(\cos \left(\dfrac{2x}{5}\right)+\sin \left(\dfrac{2x}{5}\right)\right)\left(\sin \left(\dfrac{x}{5}\right)\cos \left(\dfrac{5\pi }{12}\right)+\cos \left(\dfrac{x}{5}\right)\sin \left(\dfrac{5\pi}{12}\right)\right)$ (*)
Unfortunately, there are no ready to use expressions for both $\sin \left(\dfrac{5\pi }{12}\right)$ and $\cos \left(\dfrac{5\pi }{12}\right)$ and hence, we get stuck at this point now.
Don't give up yet, we could work on it as we see that:
$\begin{align*}\sin \left(\dfrac{5\pi }{12}\right)&=\sin \left(\dfrac{\pi }{4}+\dfrac{\pi }{6}\right)\\&=\sin \left(\dfrac{\pi }{4}\right)\cos \left(\dfrac{\pi }{6}\right)+\cos \left(\dfrac{\pi }{4}\right)\sin \left(\dfrac{\pi }{6}\right)\\&=\left(\dfrac{\sqrt{2}}{2}\dfrac{\sqrt{3}}{2}\right)+\left(\dfrac{\sqrt{2}}{2}\dfrac{1}{2}\right)\\&=\dfrac{\sqrt{6}+\sqrt{2}}{4}\end{align*}$
Similarly,
$\begin{align*}\cos \left(\dfrac{5\pi }{12}\right)&=\cos \left(\dfrac{\pi }{4}+\dfrac{\pi }{6}\right)\\&=\cos\left(\dfrac{\pi }{4}\right)\cos \left(\dfrac{\pi }{6}\right)-\sin \left(\dfrac{\pi }{4}\right)\sin \left(\dfrac{\pi }{6}\right)\\&=\left(\dfrac{\sqrt{2}}{2}\dfrac{\sqrt{3}}{2}\right)-\left(\dfrac{\sqrt{2}}{2}\dfrac{1}{2}\right)\\&=\dfrac{\sqrt{6}-\sqrt{2}}{4}\end{align*}$
But, have you considered even before you got started to work for both the exact values of $\sin \left(\dfrac{5\pi }{12}\right)$ and $\cos \left(\dfrac{5\pi }{12}\right)$ that if it is wise to continue working with what we have simplified of the given original equation?
Regardless of what your answer is, there is not quite possible to know, if our idea will work like charm or it will not contribute any significant help to solve for the problem at hand.
Alll that we could do is to stay focus, and always simplifying the expressions:
What would we get, when we replace the exact values of $\sin \left(\dfrac{5\pi }{12}\right)$ and $\cos \left(\dfrac{5\pi }{12}\right)$ into the equation (*)? We don't work it on our head, we write it all out, hopefully, we see some new perspective to proceed...
$ 2\cos \left(\dfrac{x}{5}\right)-2\sin \left(\dfrac{x}{5}\right)$
$\small =-4\left(\dfrac{\sqrt{2}}{2}\right)\left(\cos \left(\dfrac{2x}{5}\right)+\sin \left(\dfrac{2x}{5}\right)\right)\left(\sin \left(\dfrac{x}{5}\right)\cos \left(\dfrac{5\pi }{12}\right)+\cos \left(\dfrac{x}{5}\right)\sin \left(\dfrac{5\pi}{12}\right)\right)$ (*)
$\small =-4\left(\dfrac{\sqrt{2}}{2}\right)\left(\cos \left(\dfrac{2x}{5}\right)+\sin \left(\dfrac{2x}{5}\right)\right)\left(\sin \left(\dfrac{x}{5}\right) \left(\dfrac{\sqrt{6}-\sqrt{2}}{4}\right)+\cos \left(\dfrac{x}{5}\right)\left(\dfrac{\sqrt{6}+\sqrt{2}}{4}\right)\right)$
$\small =-4\left(\dfrac{\sqrt{2}}{2}\right)\left(\cos \left(\dfrac{2x}{5}\right)+\sin \left(\dfrac{2x}{5}\right)\right)\dfrac{1}{4}\left(\sin \left(\dfrac{x}{5}\right)(\sqrt{6}-\sqrt{2})+\cos \left(\dfrac{x}{5}\right)(\sqrt{6}+\sqrt{2})\right)$
$\small =-\cancel{4}\left(\dfrac{\sqrt{2}}{2}\right)\left(\cos \left(\dfrac{2x}{5}\right)+\sin \left(\dfrac{2x}{5}\right)\right)\dfrac{1}{\cancel{4}}\left(\sin \left(\dfrac{x}{5}\right)(\sqrt{6}-\sqrt{2})+\cos \left(\dfrac{x}{5}\right)(\sqrt{6}+\sqrt{2})\right)$
$\small =-\left(\dfrac{\sqrt{2}}{2}\right)\left(\cos \left(\dfrac{2x}{5}\right)+\sin \left(\dfrac{2x}{5}\right)\right)\left(\sin \left(\dfrac{x}{5}\right)(\sqrt{6}-\sqrt{2})+\cos \left(\dfrac{x}{5}\right)(\sqrt{6}+\sqrt{2})\right)$
$\small =-\left(\dfrac{\sqrt{2}}{2}\right)\left(\cos \left(\dfrac{2x}{5}\right)+\sin \left(\dfrac{2x}{5}\right)\right)\left(\sqrt{6}\left(\sin \left(\dfrac{x}{5}\right)+\cos \left(\dfrac{x}{5}\right)\right)-\sqrt{2}\left(\sin \left(\dfrac{x}{5}\right)-\cos \left(\dfrac{x}{5}\right)\right)\right)$
Will you agree with me now that we should stop right now and look for other approach?
There are just too much terms that we can't factorize them nicely at all.
Don't fret, if this method really doesn't work, then it is better off to stop continuing. There must be some good methods to solve it. Just that our attempts so far are all futile attempts.
I think the time is late now, so I will show you two credible ways of attacking this hard trigonometric equation in the next blog posts.
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