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!

- anonymous

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- anonymous

@Kainui @jim_thompson5910 @amistre64 @ganeshie8 @zepdrix @campbell_st @Loser66

- anonymous

@Luigi0210 @robtobey @sammixboo @Zarkon @pooja195 @radar

- zepdrix

Hmm the 2 is going to cause a bit of a problem :)
Not a nice friendly angle there.
But the 1 is ok.

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## More answers

- anonymous

I think the one is pi/4?

- 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

since it is for all solutions, should I add the extension? Or is there another thing I need to do?

- 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

periodic* blah

- anonymous

So I'd just say pi/4+kpi and add nEz to the right?

- zepdrix

\(\Large\rm k\in\mathbb{Z}\)
ya that works out nicely I guess :)

- anonymous

cool and I don't need to add 5pi/4+kpi? As it is another solution

- zepdrix

5pi/4 = pi/4 + 1*pi
so it's already included in our set of solutions with the k

- zepdrix

we just need to deal with this ugly tanx-2 now :d

- anonymous

lol, I simplified it to 7pi/20

- anonymous

I think that sounds right, but am not sure

- 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

I asked my teacher, but he said I needed to go further

- zepdrix

mmm ok thinking :)

- anonymous

cool

- anonymous

I found the radian in decimals and divided by pi and converted that into a fraction

- anonymous

If that is the correct process

- anonymous

and multiplied pi of course :)

- 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

yea

- anonymous

about 7/20

- anonymous

is that right? I'm not too good at this ha ha

- anonymous

I converted with some calculator

- anonymous

which gave me 7/20, which is about right

- anonymous

I rounded it for the calc

- 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

Maybe you guys have some fun technique for finding arctan(2),
maybe my brain is a lil rusty hehe

- anonymous

thanks for the help! I'll give you a medal

- zepdrix

np c:

- anonymous

:)

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