Найдите наибольший отрицательный корень уравнения cos\ pi(8х-9)/3=1/2

[latex]cos\frac{\pi (8x-9)}{3}=\frac{1}{2}\\\\\frac{\pi (8x-9)}{3}=\pm \frac{\pi}{3}+2\pi n,\; n\in Z\\\\8x-9=\pm 1+6n,\; n\in Z\\\\8x=9\pm 1+6n,\; n\in Z\\\\x=\frac{9}{8}\pm \frac{1}{8}+\frac{3n}{4}= \left \{ {{\frac{5}{4}+\frac{3n}{4}} \atop {1+\frac{3n}{4}}} \right. \\\\Pri\; \; \; n=-1:\; x= \left \{ {{\frac{1}{2}} \atop {\frac{1}{4}}} \right. \\\\Pri\; \; n=-2:\; \left \{ {{-\frac{1}{4}} \atop {-\frac{2}{4}=-\frac{1}{2}}} \right. \\\\Naibolshij\; \; otricatelnujkoren:\; \; x=-\frac{1}{4}.[/latex]