5tgx-12/13cosx-5=0 найти все корни на промежутке [4pi;11pi/2]

ОДЗ 13cosx-5≠0⇒cosx≠5/13⇒sinx≠12/13⇒tgx≠12/5 5tgx-12=0⇒tgx=12/5 x=arctg2,4+πn x=arctg2,4+5π∈заданному промежутку.

[latex]\frac{5tgx-12}{13cosx-5}=0\\\\OOF:\; cosx\ne \frac{5}{13},\; x\ne \pm arccos\frac{5}{13}+2\pi k,k\inZ[/latex] [latex]tgx=\frac{12}{5},\; x=arccos\frac{12}{5}+\pi n,\; n\in Z}\\\\Esli\; tg \alpha =\frac{12}{5},to\; cos \alpha =\frac{5}{13}\; i\; \alpha \in (0,\frac{\pi}{2})[/latex] [latex]Esli\; tg \alpha =\frac{12}{5}\; i\; \alpha \in (\pi ,\frac{3\pi}{2}),\; to\; cos \alpha \ne \frac{5}{13},\; togda\; \alpha =arctg\frac{12}{5}+\pi \\\\Otvet:\; x=arctg\frac{12}{5}+\pi +\pi n,\; n\in Z\\\\2).\; \; x=(arctg\frac{12}{5}+\pi)+4\pi\in [4\pi,\frac{11\pi}{2}][/latex] [latex]x=arctg\frac{12}{5}+5\pi \; \in [4\pi ,\frac{11\pi}{2}][/latex]