[tex]2^x+2^{-x}=\pi|\cdot 2^x\\(2^x)^2+1=\pi \cdot 2^x\\(2^x)^2-\pi \cdot 2^x+1=0\\\\t=2^x\\\\t^2-\pi t+1=0\\\Delta=(-\pi )^2-4\cdot1\cdot1=\pi^2-4\\\sqrt{\Delta}=\sqrt{\pi^2-4}\\\\t_1=\dfrac{-(-\pi)-\sqrt{\pi^2 -4}}{2\cdot1}=\dfrac{\pi-\sqrt{\pi^2 -4}}{2}\\t_2=\dfrac{-(-\pi)+\sqrt{\pi^2 -4}}{2\cdot1}=\dfrac{\pi+\sqrt{\pi^2 -4}}{2}\\\\2^x=\dfrac{\pi-\sqrt{\pi^2 -4}}{2}\\x=\log_2\dfrac{\pi-\sqrt{\pi^2 -4}}{2}=\log_2\left(\pi-\sqrt{\pi^2 -4}\right)-1[/tex]
[tex]2^x=\dfrac{\pi+\sqrt{\pi^2 -4}}{2}\\x=\log_2\dfrac{\pi+\sqrt{\pi^2 -4}}{2}=\log_2\left(\pi+\sqrt{\pi^2 -4}\right)-1\\\\[/tex]
[tex]\boxed{x\in\left\{\log_2\left(\pi-\sqrt{\pi^2 -4}\right)-1,\log_2\left(\pi+\sqrt{\pi^2 -4}\right)-1\right\}}[/tex]
