Question

tg x - 12 ctg x + 1 = 0

Answer 1

$tgx-12ctgx+1=0\\tg^2x+tgx-12=0\\D_{tgx}=1+48=49\\tgx= \left \{ -3;4 \right \} \Rightarrow x=\left \{ -arctg3+\pi k;arctg4+\pi k \right \},k\in \mathbb{Z}$