Tg[latex] \pi (8x-3)/3=- \sqrt{3} [/latex]помогите пожалуйста

[latex]tg\frac{\pi(8x-3)}{3}=-\sqrt3\iff\frac{\pi(8x-3)}{3}=-\frac{\pi}{3}+k\pi\ \ \ \ /\cdot3\\\\\pi(8x-3)=-\pi+3k\pi\ \ \ \ /:\pi\\\\8x-3=-1+3k\\\\8x=-1+3+3k\\\\8x=2+3k\ \ \ \ /:8\\\\x=\frac{2+3k}{8};\ \ \ \ (k\in\mathbb{Z})[/latex] [latex]\frac{\pi(8x-3)}{3}\neq\frac{\pi}{2}+k\pi\ \ \ \ /\cdot3\\\\\pi(8x-3)\neq\frac{3\pi}{2}+3k\pi\ \ \ \ /:\pi\\\\8x-3\neq\frac{3}{2}+3k\\\\8x\neq\frac{3}{2}+3+3k\\\\8x\neq\frac{9}{2}+3k\\\\8x\neq\frac{9+6k}{2}\ \ \ \ /:8\\\\x\neq\frac{9+6k}{16}[/latex]