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Mathematics 8 Online
OpenStudy (openstudier):

What are some solutions of the equation.

OpenStudy (openstudier):

\[\tan^2x+\tan x-2=0\]

OpenStudy (anonymous):

let y = tan x you have y^2 + y - 2 = 0 (y-2)(y+1) = 0

OpenStudy (anonymous):

to tan x = 2 or tan x = -1

OpenStudy (anonymous):

sorry it should be (y+2)(y-1) = 0

OpenStudy (anonymous):

so tan x = -2 or tan x = 1

OpenStudy (anonymous):

let \(\tan x=y\) you get \[y^2+y-2=0\\ (y+2)(y-1)=0\\ y+2=0\qquad,\qquad y-1=0 \] find y, but y=tan x so, find "x"

OpenStudy (anonymous):

lol thats what I did

OpenStudy (anonymous):

@Kanwar245 lol. yes..

OpenStudy (openstudier):

\[0<x<2\pi\]

OpenStudy (openstudier):

@electrokid and @Kanwar245

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