Ответ:

решение смотри на фотографии

[tex]\displaystyle\bf\\\frac{3^{x}+3^{-x} }{3^{x}-3^{-x} } =2\\\\\\\frac{3^{-x}\cdot(3^{2x} +1) }{3^{-x}\cdot(3^{2x}-1) }=2\\\\\\\frac{3^{2x}+1 }{3^{2x}-1 }=2\\\\\\3^{2x}=m \ \ , \ \ m > 0\\\\\\\frac{m+1}{m-1}-2=0\\\\\\\frac{m+1-2m+2}{m-1}=0\\\\\\\frac{3-m}{m-1} =0[/tex]

[tex]\displaystyle\bf\\\left[\begin{array}{ccc}3-m=0\\m-1\neq 0\end{array}\right\\\\\\\left[\begin{array}{ccc}m=3\\m\neq 1\end{array}\right\\\\\\3^{2x}=3\\\\2x=1\\\\\boxed{x=0,5}[/tex]