Решить уравнение: tg8x = tg11x

$tg(\alpha)-tg(\beta)=\frac{sin(\alpha-\beta)}{cos(\alpha)*cos(\beta)}$

$tg(8x)=tg(11x)\\\\ tg(11x)-tg(8x)=0\\\\ \frac{sin(11x-8x)}{cos(11x)*cos(8x)}=0\\\\ \frac{sin(3x)}{cos(11x)*cos(8x)}=0\\\\ \left \{ {{sin(3x)=0} \atop {cos(11x)\neq0\ \ and\ \ cos(8x)\neq0} \right. \\\\ \left \{ {{3x=\pi n,\ n\in Z} \atop {11x\neq\frac{\pi}{2}+\pi k,\ k\in Z\ \ and\ \ 8x\neq\frac{\pi}{2}+\pi p,\ p\in Z} \right. \\\\ \left \{ {{x=\frac{\pi n}{3},\ n\in Z} \atop {x\neq\frac{\pi}{22}+\frac{\pi k}{11},\ k\in Z\ \ and\ \ x\neq\frac{\pi}{16}+\frac{\pi p}{8},\ p\in Z} \right. \\\\$

$x=\frac{\pi n}{3},\ n\in Z$

Ответ: $\frac{\pi n}{3},\ n\in Z$