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**

We'll substitute (tan x)^2 = (sec x)^2 - 1

We'll re-write the equation:

(sec x)^2 - 1 = 8 - 8secx

We'll move all terms to one side:

(sec x)^2 - 1 - 8 + 8secx = 0

We'll put sec x = t

t^2 + 8t - 9 = 0

We'll write -9 = -1-8

t^2 + 8t - 8 - 1 = 0

We'll group the 1st and the last terms and the middle terms:

(t^2 - 1) + 8(t-1) = 0

We'll re-write the difference of squares:

(t-1)(t+1) + 8(t-1) = 0

(t-1)(t+1+8) = 0

(t-1)(t+9) = 0

We'll put t-1 = 0

t = 1

But t = sec x = 1/cos x

1/cos x = 1

cos x = 1

x = 2k*pi

t+9 = 0

sec x = -9

1/cos x = -9

cos x = -1/9

x = +/-arccos(1/9) + 2kpi, k is integer

**The solutions of the equation are:{2k*pi} ; {+/-arccos(1/9) + 2kpi}.**