# Solve equationtan^2x-8tanx+12=0

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We have to solve the equation: (tan x)^2 - 8 tan x + 12 = 0

let tan x = y

(tan x)^2 - 8 tan x + 12 = 0

=> y^2 - 8y + 12 = 0

=> y^2 - 6y - 2y + 12 = 0

=> y(y -6) - 2( y - 6) = 0

=> ( y - 2)(y - 6) = 0

So y = 2 and y = 6

As y = tan x

tan x = 2 and tan x = 6

=> x = arc tan 2 + n*pi and x = arc tan 6 + n*pi

Therefore the solution is

**x = arc tan 2 + n*pi and**

** x = arc tan 6 + n*pi**

To calculate the equation, it's proper to make th substitution tanx=t

(t)^2 - 8*t + 12 = 0

We'll apply quadratic formula

t1= [-(-8)+sqrt(64-48)]/2

t1=(8+4)/2

t1=6

tan x=t1

tan x=6

x = arctan 6 + k*pi

t2= (8-4)/2

t2= 2

tan x = 2

x = arctan 2 + k*pi

**The solutions of the equation are {arctan 6 + k*pi}U{arctan 2 + k*pi}.**