Question

The product of all real values of $x$ satisfying the equation
${\sin }^{-1}\cos \left(\frac{2x^2+10|x|+4}{x^2+5|x|+3}\right)=\cot \left({\cot }^{-1}\left(\frac{2-18|x|}{9|x|}\right)\right)+\frac{\pi }{2}$ is

Answer 1

${\sin }^{-1}\cos \left(\frac{2x^2+10\left|x\right|+4}{x^2+5\left|x\right|+3}\right)=\cot \left({\cot }^{-1}\left(\frac{2-18\left|x\right|}{9\left|x\right|}\right)\right)+\frac{\pi }{2}\Rightarrow -(\frac{\pi }{2}-{\sin }^{-1}\cos \left(\frac{2x^2+10\left|x\right|+4}{x^2+5\left|x\right|+3}\right))=\frac{2-18\left|x\right|}{9\left|x\right|}\Rightarrow -{\cos }^{-1}\cos \left(\frac{2x^2+10\left|x\right|+6-2}{x^2+5\left|x\right|+3}\right)=\frac{2}{9\left|x\right|}-\frac{18\left|x\right|}{9\left|x\right|}\Rightarrow -\frac{2(x^2+5\left|x\right|+3)}{(x^2+5\left|x\right|+3)}+\frac{2}{(x^2+5\left|x\right|+3)}=\frac{2}{9\left|x\right|}-2or,9\left|x\right|=x^2+5\left|x\right|+3[Assumingx\ne 0]x^2+4\left|x\right|+3=0x^2-3\left|x\right|-\left|x\right|+3=0(\left|x\right|-3)(\left|x\right|-1)=0\left|x\right|=3or,\left|x\right|=1x=3,-3\&x=1,-1$

Product of all real value$=3x-3\times |x-1|=9$