Question

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

sec^2 x-2=tan^2 x

Answer 1

\[\sec^2(x)-2=\tan^2(x)\]Is this how it's supposed to be?

@mangnagelytis

Answer 2

Remember, \(\tan^2 x + 1 = \sec^2 x\), so \(\sec^2 x - 2 = \tan x = \sec^2 - 1\)

\[\sec^2 - 2 = \sec^2 - 1\]
No solution.

Answer 3

Oops I mean sec²x − 2=sec²x − 1

Answer 4

This is all based on the assumption that what I typed is what the actual problem is. @geerky42

And in that case, there is no solution. @mangnagelytis

Answer 5

That's what I thought. Thank you!