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December 21, 2025, 03:56:54 pm

Author Topic: Dooodyo's noob questions  (Read 1756 times)  Share 

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dooodyo

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Dooodyo's noob questions
« on: April 17, 2011, 11:17:11 pm »
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Hey guys,

Just have one question how would you find all intercepts for sin 3x =1 for ( -2pie, 2pie) ?

THanks in advance

luken93

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Re: Dooodyo's noob questions
« Reply #1 on: April 17, 2011, 11:21:50 pm »
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3x = sin^-1(1)
3x = pi/2
x = pi/6

Then add/subtract period (2pi/3)

x = pi/6 - 4pi/6, pi/6, pi/6 + 4pi/6
 = -3pi/6, pi/6, 5pi/6
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Re: Dooodyo's noob questions
« Reply #2 on: April 17, 2011, 11:22:03 pm »
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sin(3x)=1 has a reference angle of pi/2
therefore: 3x=pi/2
x=pi/6     -now to get it within the domain. the period is 2pi/3, so add & minus 4pi/6 from pi/6
thus: x= -11/pi, -7pi/6, -pi/2, pi/6, 5pi/6, 3pi/2

dooodyo

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Re: Dooodyo's noob questions
« Reply #3 on: April 17, 2011, 11:25:22 pm »
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Oh haha thanks heaps Swarley and Luken

And do you always add or subtract period?

Or sometimes I thought you just add or subtract 2pie from the 1st quad angle? soooo confused  :-\

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Re: Dooodyo's noob questions
« Reply #4 on: April 17, 2011, 11:29:01 pm »
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It depends what you want to do. If you were to find solutions in a restricted domain then find the first 2 solutions, then you can add, subtract the period from/to the solutions. It just depends on what thte domain is.
If you have 23pi/4 but the questoin wants it in [0,2pi] then 13pi/6 - 2pi = pi/6.

dooodyo

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Re: Dooodyo's noob questions
« Reply #5 on: April 17, 2011, 11:31:30 pm »
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Wait so does that mean you add 2pie or add the period each time?
« Last Edit: April 18, 2011, 12:11:45 am by dooodyo »

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Re: Dooodyo's noob questions
« Reply #6 on: April 17, 2011, 11:36:09 pm »
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oh sry for the confusion. When you're finding solutions you add the period. However if you want to find an equvialent angle then you add/subtract multiples of 2pi.
I thinks that right

dooodyo

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Re: Dooodyo's noob questions
« Reply #7 on: April 25, 2011, 04:51:23 pm »
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Hey guys,

I have another noob question any help would be appreciated.

How would you find the general solution to sqrt(3)tan(pi/6-3x)-1=0 ?

why is it (pi x n)/3 and not (-pi x n)/3 ?

brightsky

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Re: Dooodyo's noob questions
« Reply #8 on: April 25, 2011, 05:09:39 pm »
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They are the same, since n is any integer.
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Re: Dooodyo's noob questions
« Reply #9 on: April 25, 2011, 05:59:31 pm »
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Hey guys,

I have another noob question any help would be appreciated.

How would you find the general solution to sqrt(3)tan(pi/6-3x)-1=0 ?

why is it (pi x n)/3 and not (-pi x n)/3 ?
They are the same, since n is any integer.

Provided your solution is followed by "n E Z"

You could, however, say +/- both those options, but you'd need to say "n E N U O"
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brightsky

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Re: Dooodyo's noob questions
« Reply #10 on: April 25, 2011, 06:04:27 pm »
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You could, however, say +/- both those options, but you'd need to say "n E N U O"

There are infinite many ways to write the 'general solution'. I'd say just write one, with n E Z at the end as Aurelian said. I'm pretty sure all the 'variations' would be accepted, as all of them are technically correct.
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Re: Dooodyo's noob questions
« Reply #11 on: April 25, 2011, 06:11:52 pm »
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Yeah, I'd stick to the n E Z variation like brightsky said - less room for silly mistakes.

Also, this might be of help;

http://vce.atarnotes.com/forum/index.php/topic,35962.msg382191.html#msg382191
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dooodyo

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Re: Dooodyo's noob questions
« Reply #12 on: April 25, 2011, 10:14:42 pm »
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wait does " n E N U O " mean element of natural no.s union 0 ?

Aurelian

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Re: Dooodyo's noob questions
« Reply #13 on: April 25, 2011, 11:16:56 pm »
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wait does " n E N U O " mean element of natural no.s union 0 ?

Yep :) I'd have used latex but I don't know how haha sorry :P
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dooodyo

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Re: Dooodyo's noob questions
« Reply #14 on: April 26, 2011, 10:40:23 am »
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Thanks heaps  :D