## Wednesday, June 17, 2015

### Smartest Way of Solving Hardest Trigonometric Equation

Solve the trigonometric equation.

$\sqrt{2} \cos \left(\dfrac{x}{5}-\dfrac{\pi }{12}\right)-\sqrt{6}\sin \left(\dfrac{x}{5}-\dfrac{\pi}{12}\right)=2\left(\sin \left((\dfrac{x}{5}-\dfrac{2\pi}{3}\right)-\sin \left((\dfrac{3x}{5}+\dfrac{\pi}{6}\right)\right)$

An excellent observation and hence sublimely insightful solution provided by one mathematicans from the England:

LHS of the equation can be simplified, as shown in previous posts that we got:

[MATH]\color{yellow}\bbox[5px,purple]{\sqrt{2} \cos \left(\dfrac{x}{5}-\dfrac{\pi }{12}\right)-\sqrt{6}\sin \left(\dfrac{x}{5}-\dfrac{\pi}{12}\right)=2\cos\dfrac{x}{5}-2\sin\dfrac{x}{5}}[/MATH]

Whereas the first term in the RHS of the equation can be simplified to get:

[MATH]\color{yellow}\bbox[5px,green]{2\sin \left(\dfrac{x}{5}-\dfrac{2\pi}{3}\right)}[/MATH]

$=2\left(\sin \dfrac{x}{5} \cos \dfrac{2\pi}{3}-\cos \dfrac{x}{5} \sin \dfrac{2\pi}{3} \right)$

$=2\left(\sin \dfrac{x}{5}\left(\dfrac{-1}{2}\right)-\cos \dfrac{x}{5}\left(\dfrac{\sqrt{3}}{2}\right) \right)$

$=-\sin \dfrac{x}{5}-\sqrt{3}\cos \dfrac{x}{5}$

Putting these together, we see that:

$\small \color{yellow}\bbox[5px,purple]{\sqrt{2} \cos \left(\dfrac{x}{5}-\dfrac{\pi }{12}\right)-\sqrt{6}\sin \left(\dfrac{x}{5}-\dfrac{\pi}{12}\right)}\color{black}=\color{yellow}\bbox[5px,green]{2\sin \left(\dfrac{x}{5}-\dfrac{2\pi}{3}\right)}\color{black}-2\sin \left(\dfrac{3x}{5}+\dfrac{\pi}{6}\right)$

$2\cos\dfrac{x}{5}-2\sin\dfrac{x}{5}=-\sin \dfrac{x}{5}-\sqrt{3}\cos \dfrac{x}{5}-2\sin \left(\dfrac{3x}{5}+\dfrac{\pi}{6}\right)$

$\sqrt{3}\cos \dfrac{x}{5}-\sin\dfrac{x}{5}=-2\cos\dfrac{x}{5}-2\sin \left(\dfrac{3x}{5}+\dfrac{\pi}{6}\right)$

$\dfrac{\sqrt{3}}{2}\cos \dfrac{x}{5}-\dfrac{1}{2}\sin\dfrac{x}{5}=-\cos\dfrac{x}{5}-\sin \left(\dfrac{3x}{5}+\dfrac{\pi}{6}\right)$

$\cos \dfrac{\pi}{6}\cos \dfrac{x}{5}-\sin\dfrac{\pi}{6}\sin\dfrac{x}{5}=-\sin\left(\dfrac{x}{5}+\dfrac{\pi}{2}\right)-\sin \left(\dfrac{3x}{5}+\dfrac{\pi}{6}\right)$

\begin{align*}\cos \left(\dfrac{x}{5}+ \dfrac{\pi}{6}\right)&=-\left(\sin\left(\dfrac{x}{5}+\dfrac{\pi}{2}\right)+\sin \left(\dfrac{3x}{5}+\dfrac{\pi}{6}\right)\right)\\&=-2\sin \left(\dfrac{2x}{5}+\dfrac{\pi}{3}\right)\cos \left(\dfrac{x}{5}-\dfrac{\pi}{6}\right)\\&=-2\left((2\sin \left(\dfrac{x}{5}+\dfrac{\pi}{6}\right)\cos \left(\dfrac{x}{5}+\dfrac{\pi}{6}\right)\cos \left(\dfrac{x}{5}-\dfrac{\pi}{6}\right)\right)\end{align*}

Factoring it we get:

[MATH]\color{black}\bbox[5px,orange]{\cos \left(\dfrac{x}{5}+ \dfrac{\pi}{6}\right)}\color{yellow}\bbox[5px,blue]{\left(4\sin \left(\dfrac{x}{5}+\dfrac{\pi}{6}\right)\cos \left(\dfrac{x}{5}-\dfrac{\pi}{6}\right)+1\right)}\color{black}=0[/MATH]

This yields two results, where either one or both of the factors must be $0$.

1.

[MATH]\color{black}\bbox[5px,orange]{\cos \left(\dfrac{x}{5}+ \dfrac{\pi}{6}\right)}\color{black}=0[/MATH] yields [MATH]x=\frac{5\pi}{3}+10n\pi[/MATH], where $n$ is integer.

2.

[MATH]\color{yellow}\bbox[5px,blue]{\left(4\sin \left(\dfrac{x}{5}+\dfrac{\pi}{6}\right)\cos \left(\dfrac{x}{5}-\dfrac{\pi}{6}\right)+1\right)}\color{black}=0[/MATH]

$4\sin \left(\dfrac{x}{5}+\dfrac{\pi}{6}\right)\cos \left(\dfrac{x}{5}-\dfrac{\pi}{6}\right)+1=0$ shows that:

$2\left(2\sin \left(\dfrac{x}{5}+\dfrac{\pi}{6}\right)\cos \left(\dfrac{x}{5}-\dfrac{\pi}{6}\right)\right)=-1$

$\sin \dfrac{2x}{5}+\sin \dfrac{\pi}{3}=-\dfrac{1}{2}$

\begin{align*}\sin \dfrac{2x}{5}&=-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\\&=-\dfrac{1+\sqrt{3}}{2}\\&\le -1\end{align*}

which means this part has no real solutions to offer.

Therefore, the answers for this problem are [MATH]x=\frac{5\pi}{3}+10n\pi[/MATH], where $n$ is integer.