Question

Help finding all solutions of this equation in the interval [0, 2pi)? (Will give medal & fan)
tan^x - sec x = -1

Answer 1

Are you sure your equation is correct as written?

Answer 2

There seems to be something missing from the tangent

Answer 3

tan^2 x - sec x = -1, sorry

Answer 4

Okay, so basically, what you have to do here is convert the trigonometric equation to a quadratic equation.

Answer 5

How would I do that?

Answer 6

Well, if you remember your Pythagorean Identities, you would know that
\(\tan^2x = \sec^2x - 1\) right?

Answer 7

Yeah

Answer 8

So since that is true, this means \(\tan^2 x - \sec x = -1\) can be re-written as

\(\sec^2x - 1 - sec x = -1\)

Answer 9

Furthermore,

If we add 1 to both sides, we'll have

\(\sec^2x - \sec x = 0\) Do you agree?

Answer 10

Yes I agree

Answer 11

And now we have the trigonometric equation in quadratic form. Do you see what we might be able to do for the next step?

Answer 12

hmm... factor...?

Answer 13

Yes, correct. So how would the equation look after factoring?

Answer 14

(sin x)(sin x) - sin x?

Answer 15

Hmmm. I thought we were working with sec(x)

Answer 16

My bad, the same thing but with sec. I'm exhausted lol

Answer 17

Well, actually, maybe we should let y = sec(x) for now.

Answer 18

If that's the case, then \(\sec^2(x) - \sec(x) = 0\) would become \(y^2 - y = 0\)

Answer 19

So factor \(y^2 - y = 0\)

Answer 20

I'm not really that great at factoring /: (y - 1)(y + 1)?

Answer 21

Sorry my internet went down for a little. I'm still confused

Answer 22

\(y^2 - y = 0\) factors to

\(y(y - 1) = 0\)

Likewise
\(\sec(x)(\sec(x) - 1) = 0\)

Furthermore

\(\sec(x) = 0\) or \(\sec(x) = 1\)