anonymous
  • anonymous
Thanks for helping! I was given a pre calc problem which I simplified to: (tanx-2)(tanx-1)=0 The question asks me to solve from there. However, it doesn't ask me to solve in terms of 0 to 2pi, it asks for all solutions. I know that I have to add an extension once I find the final solution, but am having a hard time figuring out how to do so. I know it looks something like+n*pi*k or something like that. Thank You!
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@Kainui @jim_thompson5910 @amistre64 @ganeshie8 @zepdrix @campbell_st @Loser66
anonymous
  • anonymous
@Luigi0210 @robtobey @sammixboo @Zarkon @pooja195 @radar
zepdrix
  • zepdrix
Hmm the 2 is going to cause a bit of a problem :) Not a nice friendly angle there. But the 1 is ok.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I think the one is pi/4?
zepdrix
  • zepdrix
Applying our Zero-Factor Property,\[\Large\rm \tan x-1=0\]Add 1 to each side,\[\Large\rm \tan x=1\] Ok good, yes this corresponds to pi/4 in the first quadrant.
anonymous
  • anonymous
since it is for all solutions, should I add the extension? Or is there another thing I need to do?
zepdrix
  • zepdrix
\[\Large\rm x=\frac{\pi}{4}+k \pi, \quad k=0,\pm1,\pm2,...\]Tangent is a little different than sine and cosine. It is period in `pi` not in `2pi`. So we want to allow multiples of pi to be added on and give us the same value through tangent.
zepdrix
  • zepdrix
periodic* blah
anonymous
  • anonymous
So I'd just say pi/4+kpi and add nEz to the right?
zepdrix
  • zepdrix
\(\Large\rm k\in\mathbb{Z}\) ya that works out nicely I guess :)
anonymous
  • anonymous
cool and I don't need to add 5pi/4+kpi? As it is another solution
zepdrix
  • zepdrix
5pi/4 = pi/4 + 1*pi so it's already included in our set of solutions with the k
zepdrix
  • zepdrix
we just need to deal with this ugly tanx-2 now :d
anonymous
  • anonymous
lol, I simplified it to 7pi/20
anonymous
  • anonymous
I think that sounds right, but am not sure
zepdrix
  • zepdrix
Woah I'm not really sure 0_O I was just going to be lazy and call the angle \(\Large\rm \arctan(2)\)
anonymous
  • anonymous
I asked my teacher, but he said I needed to go further
zepdrix
  • zepdrix
mmm ok thinking :)
anonymous
  • anonymous
cool
anonymous
  • anonymous
I found the radian in decimals and divided by pi and converted that into a fraction
anonymous
  • anonymous
If that is the correct process
anonymous
  • anonymous
and multiplied pi of course :)
zepdrix
  • zepdrix
arctan(2) = 1.1071487... divide by pi gives: \(\Large\rm 0.352416382\pi\) then what did you do? 0_o how did you convert to fraction?
anonymous
  • anonymous
yea
anonymous
  • anonymous
about 7/20
anonymous
  • anonymous
is that right? I'm not too good at this ha ha
anonymous
  • anonymous
I converted with some calculator
anonymous
  • anonymous
which gave me 7/20, which is about right
anonymous
  • anonymous
I rounded it for the calc
zepdrix
  • zepdrix
im not sure how to get an exact answer :d its look like it's some irrational multiple of pi, like some weird square root maybe. so ya approximating is probably the right way to go. I would leave it as a decimal though maybe. \[\Large\rm x=0.352\pi+k \pi\]\[\Large\rm x=(0.352+k)\pi,\qquad k\in\mathbb{Z}\]
zepdrix
  • zepdrix
Maybe you guys have some fun technique for finding arctan(2), maybe my brain is a lil rusty hehe
anonymous
  • anonymous
thanks for the help! I'll give you a medal
zepdrix
  • zepdrix
np c:
anonymous
  • anonymous
:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.