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anonymous
 one year ago
Thanks for helping!
I was given a pre calc problem which I simplified to: (tanx2)(tanx1)=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!
anonymous
 one year ago
Thanks for helping! I was given a pre calc problem which I simplified to: (tanx2)(tanx1)=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!

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Kainui @jim_thompson5910 @amistre64 @ganeshie8 @zepdrix @campbell_st @Loser66

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Luigi0210 @robtobey @sammixboo @Zarkon @pooja195 @radar

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Hmm the 2 is going to cause a bit of a problem :) Not a nice friendly angle there. But the 1 is ok.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think the one is pi/4?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Applying our ZeroFactor Property,\[\Large\rm \tan x1=0\]Add 1 to each side,\[\Large\rm \tan x=1\] Ok good, yes this corresponds to pi/4 in the first quadrant.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since it is for all solutions, should I add the extension? Or is there another thing I need to do?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3\[\Large\rm x=\frac{\pi}{4}+k \pi, \quad k=0,\pm1,\pm2,...\]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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I'd just say pi/4+kpi and add nEz to the right?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3\(\Large\rm k\in\mathbb{Z}\) ya that works out nicely I guess :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cool and I don't need to add 5pi/4+kpi? As it is another solution

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.35pi/4 = pi/4 + 1*pi so it's already included in our set of solutions with the k

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3we just need to deal with this ugly tanx2 now :d

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol, I simplified it to 7pi/20

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think that sounds right, but am not sure

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Woah I'm not really sure 0_O I was just going to be lazy and call the angle \(\Large\rm \arctan(2)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I asked my teacher, but he said I needed to go further

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I found the radian in decimals and divided by pi and converted that into a fraction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If that is the correct process

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and multiplied pi of course :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3arctan(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
 one year ago
Best ResponseYou've already chosen the best response.0is that right? I'm not too good at this ha ha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I converted with some calculator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which gave me 7/20, which is about right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I rounded it for the calc

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3im 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
 one year ago
Best ResponseYou've already chosen the best response.3Maybe you guys have some fun technique for finding arctan(2), maybe my brain is a lil rusty hehe

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks for the help! I'll give you a medal
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