12cosx-5sinx=-6,5 help me please

Решите задачу:

$12cosx-5sinx=-6,5\; |:13\; \; (12^2+5^2=169=13^2)\\\\ \frac{12}{13}\cdot cosx- \frac{5}{13}\cdot sinx=- \frac{1}{2} \\\\ \frac{12}{13}=cos \beta \; ,\; \; \frac{5}{13}=sin \beta \; \; \; (sin^2 \beta +cos^2\beta =1)\\\\tg \beta =\frac{5}{12} \; \; \to \; \; \beta =arctg\frac{5}{12}\\\\cos \beta \cdot cosx-sin \beta \cdot sinx=-\frac{1}{2}\\\\cos(x+ \beta )=-\frac{1}{2}\\\\x+ \beta =\pm (\pi -\frac{\pi}{3})+2\pi n=\pm \frac{2\pi }{3}+2\pi n,\; n\in Z\\\\x=\pm \frac{2\pi}{3}-\beta +2\pi n,\; n\in Z$

$x=\pm \frac{2\pi}{3}-arctg\frac{5}{12}+2\pi n,\; n\in Z$