Question

помогите решить!!!!!!​

Answer 1

$tgx+ctg\frac{7\pi}{6}=tg\frac{7\pi}{8}-\sqrt3\cdot tg\frac{\pi}{8}\cdot tgx\\\\\star \; \; tg\frac{7\pi}{8}=tg(\pi-\frac{\pi}{8})=-tg\frac{\pi}{8}\\\\\star \; \; ctg\frac{7\pi}{6}=ctg(\pi-\frac{\pi}{6})=-ctg\frac{\pi}{6}=-\sqrt3\\\\\\tgx-\sqrt3=-tg\frac{\pi}{8}-\sqrt3\cdot tg\frac{\pi}{8}\cdot tgx$

$tgx+\sqrt3\cdot tg\frac{\pi}{8}\cdot tgx=-tg\frac{\pi}{8}+\sqrt3\\\\tgx\cdot (1+\sqrt3\cdot tg\frac{\pi}{8})=\sqrt3-tg\frac{\pi}{8}\\\\tgx=\frac{\sqrt3-tg\frac{\pi}{8}}{1+\sqrt3\cdot tg\frac{\pi}{8}}\; \; ,\; \; \; \; \sqrt3=tg\frac{\pi}{3}\\\\tgx=\frac{tg\frac{\pi}{3}-tg\frac{\pi}{8}}{1+tg\frac{\pi}{3}\cdot \, tg\frac{\pi}{8}}\; \; \to \; \; \; tgx=tg(\frac{\pi}{3}-\frac{\pi}{8})\\\\tgx=tg\frac{5\pi }{24}\\\\x=arctg(tg\frac{5\pi}{24})+\pi n\; ,\; n\in Z\; ,\; -\frac{\pi}{2}<\frac{5\pi}{24}<\frac{\pi}{2}$

$\boxed {\; x=\frac{5\pi}{24}+\pi n\; ,\; n\in Z\; }$

P.S.  В условии в левой части скорее всего знак (+), а не (-), т.к. с (-) формулы  tg(a-b) или  tg(a+b) не будет получаться.