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I think it is A and B. But this is me taking a guess. I am terrible at trig. Sorry

Yes. I am thinking you factor to get tan^2x = -2tan(x)+1?

no factor as you would any quadratic
like
x^2 + 3x + 2
= (x + 1)(x + 2)

oh sorry - on second look this will not factor
so you need to use the quadratic formula

So would a = tan^2x b =2tanx and c=-1?

no a , b and c are the coefficients of tan^2x and tan x
so a = 1 , b = 2 and c = -1

So x = +/- sqrt (2) in the end right?

*-1 +/- sqrt(2)

-2 +/- sqrt(4 - 4*1*-1) / 2
= -2 +/- sqrt8 / 2
= (-2 +/- 2 * sqrt2 )/ 2
= -1 +/- sqrt2
yep

Awesome! So now do I just plug in this to my original equation?

eg angle with tangent 0.4142 = 0.39269 radians
now check if pi/8 gives the same value

Yes it does!

So I'm assuming that is one

ok so thats one result
pi/8 + npi

now try the other value for tan x

-1 - sqrt 2 = -2.4142 degrees right?

yes

Ok so then the radians of -2.4142 is -0.042135

I don't think I did it right

no the -2.4142 is the value of tan x
so you need tan-1(-2.4242)

Oh ok that would make it -1.1781 which is basically the same as 3pi/8 right?

So the answers would be pi/8 + npi and 3pi/8 + npi

yes

Great!! Thank you for taking the time to help me! I feel I really understand now! Thank you!

Weird... I hit the sumit button and it told me it was 5pi/8 instead of 3/pi/8

lets check that out

\[\frac{2tanx}{1-\tan^2x} = 1 \\ \tan(2x)=1\]

\[\tan(u)=1 \text{ when } u=?\]

ahh
i forgot that identity !

u = pi/4 of course
and x will be pi/8

much easier that way lol!

and there is also the u=5pi/4
so yep x=5pi/8

yes

-1.1781 is an incorrect result then?
- an extraneous result right?

does give tan(2x) approx 1
what number are you approximating ?

oh -3pi/8 is what you are approximating

sorry I dont follow you
-yes

that gives 1

when you plug in

and -3pi/8 is equivalent to 5pi/8

no - i was wrong its not extraneous

still i think lehmad learned something along the line...

Yes I learned a lot! Thank you guys for the help.