What is the angle that verifies identity tan^2x=6sec x-10?

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

We'll re-write the equation moving all terms to the left
side:

(sec x)^2 - 1 - 6sec x + 10 =
0

We'll combine like
terms:

(sec x)^2  - 6sec x + 9 =
0

We notice that the expression is a perfect
square:

(sec x - 3)^2 =
0

We'll put sec x - 3 = 0

sec
x = 3

But sec x = 1/cos x => 1/cos x  =
3

cos x = 1/3

Since the value
of cosine is in the interval [-1 ; 1], that means that it doesexist an angle that
verifies the identity.

We'll determine the angle using
inverse trigonometric function.

x = +/- arccos (1/3) +
2kpi

All the angles x that verify the
equation are {+/- arccos (1/3) + 2kpi}.