Question

решите уравнеие ctgx-4=5tng​

Answer 1

$\displaystyle ctgx-4=5tgx\\\\\text{OD3} :\quad \left[\begin{array}{ccc}sinx\neq0\\cosx\neq0\end{array}\right\quad \rightarrow\quad \left[\begin{array}{ccc}x\neq\pi n;\,\,n\in Z\\x\neq\frac{\pi}2+\pi n;\,\,n\in Z\end{array}\right\quad \rightarrow\quad x\neq\frac{\pi n}2;\,\,n\in Z\\\\\\\frac{1}{tgx}-4=5tgx\\\\1-4tgx=5tg^2x\\\\5tg^2x+4tgx-1=0\\\\tgx=t,\quad t\in R\\\\5t^2+4t-1=0\\\\\text{D}=16+20=36=6^2\\\\t_1=\frac{-4+6}{10}=\frac{2}{10}=\frac{1}5\\\\t_2=\frac{-4-6}{10}=-\frac{10}{10}=-1$

$\displaystyle \left[\begin{array}{ccc}\displaystyle tgx=\frac{1}5\\\\tgx=-1\end{array}\right\quad\rightarrow\quad\boxed{\left[\begin{array}{ccc}\displaystyle x=arctg\bigg(\frac{1}5\bigg)+\pi n;\,\,n\in Z\\\\\displaystyle x=-\frac{\pi}4+\pi n;\,\,n\in Z\end{array}\right}$