\(\cos \dfrac{{16\pi }}{{16{x^2}-8x + 49}} = \dfrac{1}{{{\rm{t}}{{\rm{g}}^2}\pi x + {\rm{ct}}{{\rm{g}}^2}\pi x}}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\cos \dfrac{{16\pi }}{{16{x^2}-8x + 49}} = \dfrac{1}{{{\rm{t}}{{\rm{g}}^2}\pi x + \dfrac{1}{{{\rm{t}}{{\rm{g}}^2}\pi x}}}}.\)
Воспользуемся тем, что \(f\left( x \right) + \dfrac{1}{{f\left( x \right)}} \ge 2\), если \(f\left( x \right) > 0\). Поэтому:
\({\rm{t}}{{\rm{g}}^2}\pi x + \dfrac{1}{{{\rm{t}}{{\rm{g}}^2}\pi x}} \ge 2;\,\,\,\,\,\,\,\,\,0 < \dfrac{1}{{{\rm{t}}{{\rm{g}}^2}\pi x + \dfrac{1}{{{\rm{t}}{{\rm{g}}^2}\pi x}}}} \le \dfrac{1}{2}.\)
Рассмотрим левую часть:
\(16{x^2}-8x + 49 = {\left( {4x-1} \right)^2} + 48 \ge 48;\,\,\,\,\,\,0 < \dfrac{{16\pi }}{{{{\left( {4x-1} \right)}^2} + 48}} \le \dfrac{\pi }{3};\,\,\,\,\,\,\dfrac{1}{2} \le \cos \dfrac{{16\pi }}{{{{\left( {4x-1} \right)}^2} + 48}} < 1.\)
Поэтому уравнение будет иметь решение, если:
\(\left\{ {\begin{array}{*{20}{c}}{\cos \dfrac{{16\pi }}{{16{x^2}-8x + 49}} = \dfrac{1}{2},}\\{{\rm{t}}{{\rm{g}}^2}\pi x + \dfrac{1}{{{\rm{t}}{{\rm{g}}^2}\pi x}} = 2\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\cos \dfrac{{16\pi }}{{{{\left( {4x-1} \right)}^2} + 48}} = \dfrac{1}{2},\\{\rm{tg}}\,\pi x = \pm 1\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\cos \dfrac{{16\pi }}{{{{\left( {4x-1} \right)}^2} + 48}} = \dfrac{1}{2},\\x = \dfrac{1}{4} + \dfrac{k}{2},\,\,\,\,k \in Z\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = \dfrac{1}{4}.\)
Ответ: \(\dfrac{1}{4}.\)