MHB Solve Trig Challenge: Find All Values of x

[anemone]

Find all values of $x$ which satisfy $\tan \left( x+\dfrac{\pi}{18}\right)\tan \left( x+\dfrac{\pi}{9}\right)\tan \left( x+\dfrac{\pi}{6}\right)=\tan x$.

 

[kaliprasad]

Spoiler

$\tan\, x \cot (x +\frac{\pi}{18}) = \tan (x+\frac{\pi}{9}) \tan (x + \frac{\pi}{6})$

or $\dfrac{\sin\, x\ cos (x+ \frac{\pi}{18})}{cos\, x \,\sin (x+\frac{\pi}{18})} = \dfrac{sin (x+ \frac{\pi}{9})sin (x + \frac{\pi}{6})}{\cos(x+ \frac{\pi}{9}) \cos (x + \frac{\pi}{6})}$

using compnendo and dividendo we have

$\dfrac{\sin\, x \cos (x+ \frac{\pi}{18}) +cos\, x sin (x+\frac{\pi}{18})}{\cos\, x sin (x+\frac{\pi}{18}) - \sin\, x \cos (x+ \frac{\pi}{18})}$ =
$\dfrac{\cos(x+ \frac{\pi}{9}) cos (x + \frac{\pi}{6})+ \sin (x+\frac{\pi}{9}) \sin (x +\frac{\pi}{6} )}{\cos(x+ \frac{\pi}{9})\ cos (x + \frac{\pi}{6})- \sin (x+ \frac{\pi}{9}) sin (x + \frac{\pi}{6})}$
or $\dfrac{sin (2x + \frac{\pi}{18})}{sin(\frac{\pi}{18})} =\dfrac {cos \frac{\pi}{18}}{cos (2x + \frac{5\pi}{18})}$

or $\sin (2x+\frac{\pi}{18})\ cos (2x + \frac{5\pi}{18}) = \sin \frac{\pi}{18}\cos\frac{\pi}{18}$
or $2\sin (2x+\frac{\pi}{18})\ cos (2x + \frac{5\pi}{18}) =2 \sin \frac{\pi}{18}\cos\frac{\pi}{18}$

or $\sin (4x + \frac{\pi}{3}) – \sin \frac{2\pi}{9} = 2 \sin \frac{\pi}{9}$
or or $\sin (4x + \frac{\pi}{3})= \sin \frac{\pi}{9} + sin \frac{2\pi}{9} = 2 \sin (\frac{\frac{\pi}{9} +\frac{ 2pi}{9}}{2}) \cos (\frac{\frac{\pi}{9} -\frac{ 2pi}{9}}{2} ) = \cos\frac{\pi}{18} $
or $\cos (\frac{\pi}{6}-4x)= \cos \frac{\pi}{18}$

or $\cos (4x – \frac{\pi}{6}) = \cos \frac{\pi}{18}$

or $4x – \frac{pi}{6} = 2n\pi \pm \frac{\pi}{18}$

or $x = \frac{2n\pi + \frac{\pi}{6}\pm \frac{\ pi}{18}}{4}$

 

[anemone]

Spoiler

$\tan\, x \cot (x +\frac{\pi}{18}) = \tan (x+\frac{\pi}{9}) \tan (x + \frac{\pi}{6})$

or $\dfrac{\sin\, x\ cos (x+ \frac{\pi}{18})}{cos\, x \,\sin (x+\frac{\pi}{18})} = \dfrac{sin (x+ \frac{\pi}{9})sin (x + \frac{\pi}{6})}{\cos(x+ \frac{\pi}{9}) \cos (x + \frac{\pi}{6})}$

using compnendo and dividendo we have

$\dfrac{\sin\, x \cos (x+ \frac{\pi}{18}) +cos\, x sin (x+\frac{\pi}{18})}{\cos\, x sin (x+\frac{\pi}{18}) - \sin\, x \cos (x+ \frac{\pi}{18})}$ =
$\dfrac{\cos(x+ \frac{\pi}{9}) cos (x + \frac{\pi}{6})+ \sin (x+\frac{\pi}{9}) \sin (x +\frac{\pi}{6} )}{\cos(x+ \frac{\pi}{9})\ cos (x + \frac{\pi}{6})- \sin (x+ \frac{\pi}{9}) sin (x + \frac{\pi}{6})}$
or $\dfrac{sin (2x + \frac{\pi}{18})}{sin(\frac{\pi}{18})} =\dfrac {cos \frac{\pi}{18}}{cos (2x + \frac{5\pi}{18})}$

or $\sin (2x+\frac{\pi}{18})\ cos (2x + \frac{5\pi}{18}) = \sin \frac{\pi}{18}\cos\frac{\pi}{18}$
or $2\sin (2x+\frac{\pi}{18})\ cos (2x + \frac{5\pi}{18}) =2 \sin \frac{\pi}{18}\cos\frac{\pi}{18}$

or $\sin (4x + \frac{\pi}{3}) – \sin \frac{2\pi}{9} = 2 \sin \frac{\pi}{9}$
or or $\sin (4x + \frac{\pi}{3})= \sin \frac{\pi}{9} + sin \frac{2\pi}{9} = 2 \sin (\frac{\frac{\pi}{9} +\frac{ 2pi}{9}}{2}) \cos (\frac{\frac{\pi}{9} -\frac{ 2pi}{9}}{2} ) = \cos\frac{\pi}{18} $
or $\cos (\frac{\pi}{6}-4x)= \cos \frac{\pi}{18}$

or $\cos (4x – \frac{\pi}{6}) = \cos \frac{\pi}{18}$

or $4x – \frac{pi}{6} = 2n\pi \pm \frac{\pi}{18}$

or $x = \frac{2n\pi + \frac{\pi}{6}\pm \frac{\ pi}{18}}{4}$

Thanks, kaliprasad for your solution using the method that has been proven to be such an useful algebraic way of simplification! (Sun)