Question

решите уравнение
- tgx+ 24ctg = 2

Answer 1

u = tgx

-u + 24/u = 2

u^2 + 2u - 24 = 0

(u - 4)(u + 6) = 0

tgx = u = {-6, 4}

x = {-arctg6 + Пk, arctg4 + Пk}

Answer 2

$-tgx+24\, ctgx=2\\\\-tgx+\frac{24}{tgx}-2=0\\\\\frac{-tg^2x-2\, tgx+24}{tgx}=0\; ,\; \; tgx\ne 0\; \; \to \; \; x\ne \pi n,\; n\in Z\\\\tg^2x+2\, tgx-24=0\\\\t=tgx\; ,\; \; t^2+2t-24=0\; ,\; \; D/4=1+24=25\; ,\; t_{1,2}=-1\pm 5\\\\t_1=-6\; ,\; \; t_2=4\\\\a)tgx=-6\; ,\; \; \underline {x=-arctg6+\pi n,\; n\in Z}\\\\b)\; \; tgx=4\; ,\; \; \underline {x=arctg4+\pi k,\; k\in Z}$