\(\cos \dfrac{{\pi \left( {8x + 1} \right)}}{6} = \dfrac{{\sqrt 3 }}{2}\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{\pi \left( {8x + 1} \right)}}{6} = \pm \dfrac{\pi }{6} + 2\pi n\left| { \cdot 6\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\pi \left( {8x + 1} \right) = \pm \pi + 12\pi n\left| {:\pi \,\,\,\,\, \Leftrightarrow } \right.} \right.\)
\( \Leftrightarrow \,\,\,\,\,8x + 1 = \pm 1 + 12n\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{8x = 12n\,\,\,\,\,\,\,\,\,\,\,\,}\\{8x = -2 + 12n}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{{3n}}{2};\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = -\dfrac{1}{4} + \dfrac{{3n}}{2},\,\,\,\,n\, \in \,Z.}\end{array}} \right.\)
Рассмотрим \(x = \dfrac{{3n}}{2},\,\,n\, \in \,Z\). Если \(n = 0\), то \(x = 0\); если \(n = 1\), то \(x = 1,5\).
Рассмотрим \(x = -\dfrac{1}{4} + \dfrac{{3n}}{2},\,\,n\, \in \,Z\). Если \(n = 0\), то \(x = -0,25\); если \(n = 1\), то \(x = 1,25\).
Следовательно, наименьший положительный корень \(x = 1,25\).
Ответ: 1,25.