[tex]\displaystyle 3^x+10*\frac{3^3}{3^x}\geq 37\\\\3^x=t; t > 0\\\\ t+270\frac{1}{t}-37\geq 0\\\\\frac{t^2-37t+270}{t}\geq 0[/tex]

решим методом интервалов

[tex]\displaystyle \left \{ {{t^2-37t+270=0} \atop {t\neq 0; t > 0}} \right. \\\\\\t^2-37t+270=0\\\\D=1369-1080=289\\\\t_{1.2}=\frac{37 \pm 17}{2}\\\\t_1=10; t_2=27[/tex]

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[tex]\displaystyle 0 < t\leq 10; t\geq 27\\\\3^x\leq 10; 3^x\geq 27\\\\x\leq log_310; x\geq 3[/tex]

Ответ: [tex]\displaystyle (-oo; log_310] [3;+oo)[/tex]