Trigonometric exercise mixed with functions asked by Dan Sitaru

Exercise:

Let $f,g:\left( 0,\infty  \right)\to \mathbb{R}$ be two functions defined to be

           $f\left( x \right)=\sin \left( \ln \frac{x\left( x+1 \right)}{x+2} \right)+\sin \left( \ln \frac{\left( x+1 \right)\left( x+2 \right)}{x} \right)$ and

          $g\left( x \right)=\sin \left( \ln \left( x\left( x+1 \right)\left( x+2 \right) \right) \right)-\sin \left( \ln \frac{x\left( x+2 \right)}{x+1} \right)$

Solve , $f\left( x \right)=g\left( x \right)$

Solution: we know that $\sin a\pm \sin b=2\sin \left( \frac{a\pm b}{2} \right)\cos \left( \frac{a\mp b}{2} \right)$

$f\left( x \right)=2\sin \left( \frac{\ln \frac{x\left( x+1 \right)}{x+2}+\ln \frac{\left( x+1 \right)\left( x+2 \right)}{x}}{2} \right)\cos \left( \frac{\ln \frac{x\left( x+1 \right)}{x+2}-\ln \frac{\left( x+1 \right)\left( x+2 \right)}{x}}{2} \right)$

       $=2\sin \left( \frac{\ln \left( \frac{x{{\left( x+1 \right)}^{2}}\left( x+2 \right)}{x\left( x+2 \right)} \right)}{2} \right)\cos \left( \frac{\ln \left( \frac{x\left( x+1 \right)}{x+2}\times \frac{x}{\left( x+1 \right)\left( x+2 \right)} \right)}{2} \right)$

        $=2\sin \left( \ln \left( x+1 \right) \right)\cos \left( \frac{1}{2}\ln \left( \frac{{{x}^{2}}}{{{\left( x+2 \right)}^{2}}} \right) \right)=2\sin \left( \ln \left( x+1 \right) \right)\cos \left( \ln \frac{x}{x+2} \right)$

$g\left( x \right)=2\cos \left( \frac{\ln \left( x\left( x+1 \right)\left( x+2 \right) \right)+\ln \left( \frac{x\left( x+2 \right)}{x+1} \right)}{2} \right)\sin \left( \frac{\ln \left( x\left( x+1 \right)\left( x+2 \right) \right)-\ln \left( \frac{x\left( x+2 \right)}{x+1} \right)}{2} \right)$

        $=2\cos \left( \ln \left( x\left( x+2 \right) \right) \right)\sin \left( \frac{1}{2}\ln \left( \frac{x\left( x+1 \right)\left( x+2 \right)}{{}}\times \frac{x+1}{x\left( x+2 \right)} \right) \right)$

       $=2\cos \left( \ln \left( x\left( x+2 \right) \right) \right)\sin \left( \ln \left( x+1 \right) \right)$

Now $f\left( x \right)=g\left( x \right)\Leftrightarrow 2\sin \left( \ln \left( x+1 \right) \right)\cos \left( \ln \frac{x}{x+2} \right)=2\cos \left( \ln \left( x\left( x+2 \right) \right) \right)\sin \left( \ln \left( x+1 \right) \right)$

$\Leftrightarrow \sin \left( \ln \left( x+1 \right) \right)\left[ \cos \left( \ln \frac{x}{x+2} \right)-\cos \left( \ln \left( x\left( x+2 \right) \right) \right) \right]=0$

$\Leftrightarrow \sin \left( \ln \left( x+1 \right) \right)=0\,\,\,\,or\,\,\,\cos \left( \ln \frac{x}{x+2} \right)=\cos \left( \ln \left( x\left( x+2 \right) \right) \right)$

$\Leftrightarrow \sin \left( \ln \left( x+1 \right) \right)=\sin 0\,\,or\,\,\,\cos \left( \ln \frac{x}{x+2} \right)=\cos \left( \ln x\left( x+2 \right) \right)$

So Case 1:: $\ln \left( x+1 \right)=0+2k\pi \,\,or\,\,\ln \left( x+1 \right)=\pi +2k\pi \,=\pi \left( 2k+1 \right)$

$\Rightarrow x={{e}^{2k\pi }}-1$ or $x={{e}^{\pi \left( 2k+1 \right)}}-1$ ; $k=0,1,...$

Case 2:: $\ln \frac{x}{x+2}=\ln x\left( x+2 \right)\,+2k\pi \,\,or\,\,\ln \frac{x}{x+2}=-\ln \left( x\left( x+2 \right) \right)+2k\pi $

$\Rightarrow \ln \left( \frac{x}{x+2}\times \frac{1}{x\left( x+2 \right)} \right)=2k\pi \,\,\,\,or\,\,\,\ln \left( \frac{x\left( x\left( x+2 \right) \right)}{x+2} \right)=2k\pi $

$\Rightarrow \ln \left( \frac{1}{x+2} \right)=2k\pi \,\,\,\,or\,\,\,\,\ln x=k\pi \Leftrightarrow \frac{1}{x+2}={{e}^{k\pi }}\,\,or\,\,x={{e}^{k\pi }}$

Hence $x{{e}^{k\pi }}+2{{e}^{k\pi }}=1\Leftrightarrow x=\frac{1-2{{e}^{k\pi }}}{{{e}^{k\pi }}}={{e}^{-k\pi }}-2\,\,\,\,\,or\,\,\,x={{e}^{k\pi }}$ ,  $k=0,1,2,....$