Question

Find all solutions in the interval [0, 2π).

sec2x - 2 = tan2x

Answer 1

(sec2x - 2 = tan2x) or (sec^2 x - 2 = tan^2 x) ?

Answer 2

\[\sec^2x-2 +\tan^2x\]

Answer 3

= 0 right ?

Answer 4

\[\sec^2x−2=\tan^2x\]
sorry

Answer 5

looks your equation still wrong
if i use the identity :
sec^2 x = tan^2 x + 1

your equation can be
sec^2 x - 2 = tan^2 x
tan^2 x + 1 - 2 = tan^2 x
-1 = 0
nah, recheck again ur equ, please

Answer 6

sec^2 x - 2 = tan^2 x

Answer 7

yeah in the case where you have a contradiction ...1=0 the answer is no solution
use @RadEn explanation