Помогите,пожалуйста:* Буду очень благодарна,плиз*-* 7tgx-8ctgx+10=0

[latex] 7\tan x-8\cot x+10=0\qquad(D:x\noy =k\pi \wedge x\not=\dfrac{\pi}{2}+k\pi, k\in\mathbb{Z}) \\ \tan x(7-8\cot^2 x+10\cot x)=0\\ -\tan x(8\cot^2 x-10\cot x-7)=0\\ -\tan x(8\cot^2 x+4\cot x-14\cot x-7)=0\\ -\tan x(4\cot x(2\cot x+1)-7(2\cot x+1)=0\\ -\tan x(2\cot x+1)(4\cot x-1)=0\\\\ -\tan x=0\\ \tan x=0\\ x=k\pi\not \in D\\\\ 2\cot x+1=0\\ 2\cot x =-1\\ \cot x=-\dfrac{1}{2}\\ x=\cot^{-1}\left(-\dfrac{1}{2}\right)+k\pi \\\\ 4\cot x-1=0\\ 4\cot x=1\\ \cot x=\dfrac{1}{4}[/latex] [latex]x=\cot^{-1}\left(\dfrac{1}{4}\right)+k\pi\\\\ \boxed{x=\cot^{-1}\left(-\dfrac{1}{2}\right)+k\pi \vee x=\cot^{-1}\left(\dfrac{1}{4}\right)+k\pi, k\in \mathbb{Z}}[/latex]