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

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**

**Solve the equation tan^2x=8-8secx.**

We have to solve (tan x)^2 = 8 - 8*sec x.

(tan x)^2 = 8 - 8*sec x

=> (sin x)^2 / (cos x)^2 = 8 - 8/(cos x)

=> (1 - (cos x)^2) / (cos x)^2 = (8*(cos x)^2 - 8* cos x)/ (cos x)^2

=> (1 - (cos x)^2) = (8*(cos x)^2 - 8* cos x)

=> 9(cos x)^2 - 8 cos x - 1 = 0

=> 9(cos x)^2 - 9 cos x + cos x - 1 = 0

=> 9(cos x)(cos x - 1) + 1( cos x - 1) = 0

=> (9(cos x) - 1)(cos x - 1) = 0

cos x = 1 and cos x = 1/9

x = arc cos 1 and x = arc cos (1/9) and - arc cos (1/9)

**The required values are : x = 2*n*pi and x = arc cos (1/9) + 2*n*pi
and x = -arc cos (1/9) + 2*n*pi**

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