Question

How do you solve 3sec^(2)x-2tan^(2)x-4=0?

`3sec^2x-2tan^2x-4=0`

Answer 1

By using some trigonometric identities.
If you wish to see only the solutions, scroll down until the end.

Explanation:

Let's factor out the `-2` :

`3sec^2 x - 2(tan^2 x + 2) = 0`.

We can rewrite `tan^2 x + 2` as `tan^2 x + 1 + 1`.
Now, `tan^2 x + 1` is equal to `sec^2 x`, as proved below :

`tan^2 x + 1 = sin^2 x/cos^2 x + 1`.

Instead of `1`, let's use `cos^2 x/cos^2 x`.

`tan^2 x + 1 = (sin^2 x + cos^2x)/cos^2 x`

`tan^2 x + 1 = 1/cos^2 x = sec^2 x`.

Back to the original expression;

`3sec^2 x - 2(sec^2 x + 1) = 0`

`3sec^2 x - 2 sec^2 x - 2 = 0`

`sec^2 x = 2`

`1/cos x = sqrt(2)`

`cos x = 1/sqrt(2)`.

The first value of `x` that comes to mind is `pi/4`, however, that relation is true for all numbers of the form :

`2npi+pi/4`
and
`2npi-pi/4`

where `n` is an integer.

In conclusion (or tl;dr) :

The set of solutions is

`S` = `{2npi+pi/4 | n -> "Z"} uu {2npi - pi/4 | n -> "Z"}`.