find the angle x if tan^2 x-4sec x+5=0?

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

We'll re-write the
equation:

(sec x)^2 - 1 - 4sec x + 5 =
0

We'll combine like
terms:

(sec x)^2  - 4sec x + 4 =
0

We notice that the expression is a perfect
square:

(sec x - 2)^2 =
0

We'll put sec x - 2 = 0

sec
x = 2

But sec x = 1/cos x => 1/cos x  =
2

cos x = 1/2

x = +/- arccos
(1/2) + 2kpi

x = +/- (pi/3) +
2kpi

The angles x that verify the equation
are {+/- (pi/3) + 2kpi}.