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How do you solve for -180`@`  `<=` `theta` `<=` 180`@`   2sin^2 3`theta` = 1/2

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tamanimat | Student, Grade 11 | Salutatorian

Posted August 31, 2013 at 1:11 AM via web

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How do you solve for -180`@`  `<=` `theta` `<=` 180`@`   2sin^2 3`theta` = 1/2

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aruv | High School Teacher | Valedictorian

Posted August 31, 2013 at 1:34 AM (Answer #1)

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Given

`2sin^2(3theta)=1/2`

`sin^2(3theta)=1/4`

`sin(3theta)=+-(1/2)`

`sin(3theta)=+sin(pi/6)`

`` `3theta=npi+(-1)^n(pi/6)`

`theta=(npi)/3+(-1)^n(pi/18)`

`n=0`

Then `theta=pi/18`

n=1

`theta=pi/3-pi/18=(5pi)/18`

n=2

`theta=(2pi)/3+pi/18=(13pi)/18`

n=3

`theta=pi-pi/18=(17pi)/18`

Thus

`theta=pi/18,(5pi)/18,(13pi)/18 and (17pi)/18`

If

`sin(3theta)=-1/2=sin(-pi/6)`

`3theta=npi+(-1)^n(-pi/6)`

`theta=(npi)/3-(-1)^n(pi/18)`

If n=0

`theta=-pi/18`

n=-1

`theta=-pi/3+pi/18=(-5pi)/18`

n=-2

`theta=(-2pi)/3-pi/18=(-13pi)/18`

n=-3

`theta=-pi+pi/18=-(17pi)/18`

Thus

`theta=-pi/18,-(5pi)/18,-(13pi)/18 and -(17pi)/18`

So finally we can say

`theta=+-pi/18,+-(5pi)/18,+-(13pi)/18 and +-(17pi)/18`

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