Question

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7tgx-8ctgx+10=0

Answer 1

$7tan x-8cot x+10=0qquad(D:xnoy =kpi wedge xnot=dfrac{pi}{2}+kpi, kinmathbb{Z}) \ tan x(7-8cot^2 x+10cot x)=0\ -tan x(8cot^2 x-10cot x-7)=0\ -tan x(8cot^2 x+4cot x-14cot x-7)=0\ -tan x(4cot x(2cot x+1)-7(2cot x+1)=0\ -tan x(2cot x+1)(4cot x-1)=0\\ -tan x=0\ tan x=0\ x=kpinot in D\\ 2cot x+1=0\ 2cot x =-1\ cot x=-dfrac{1}{2}\ x=cot^{-1}left(-dfrac{1}{2}right)+kpi \\ 4cot x-1=0\ 4cot x=1\ cot x=dfrac{1}{4}$
$x=cot^{-1}left(dfrac{1}{4}right)+kpi\\ boxed{x=cot^{-1}left(-dfrac{1}{2}right)+kpi vee x=cot^{-1}left(dfrac{1}{4}right)+kpi, kin mathbb{Z}}$