Will Medal! If 2tanx/ (1-tan^2x) = 1, then x can equal ____. Check all that apply. A) x=5pi/8 + npi B) x=pi/8 + npi C) x=3pi/8 + npi D) x=7pi/8 + npi

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Will Medal! If 2tanx/ (1-tan^2x) = 1, then x can equal ____. Check all that apply. A) x=5pi/8 + npi B) x=pi/8 + npi C) x=3pi/8 + npi D) x=7pi/8 + npi

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I think it is A and B. But this is me taking a guess. I am terrible at trig. Sorry
well first cross multiply to give 2 tanx = 1 - tan^2 x now rearrange this quadratic: tan^2 x + 2tanx - 1 = 0 can you factor this?
Yes. I am thinking you factor to get tan^2x = -2tan(x)+1?

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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?
well they have the answer options in terms of pi so convert everything to decimals so for example -1 + sqrt2 = 0.4142 so find the angle in radians which is the tangent of 0.4142 and see if one of the options gives the the same angle
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
is that not included in the solution: \[\frac{5\pi}{8}+n \pi \\ \text{ if } n=-1 \text{ then } \frac{5\pi}{8}-\pi=\frac{-3\pi}{8}\] so I'm not sure why you say it is extraneous
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.

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