Solve equation tan^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
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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}.
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