mathslover
  • mathslover
find the general solution for : \(\sin^2 \theta -2 \cos \theta + \frac{1}{4} = 0 \)
Trigonometry
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[Sin^2\theta = 1-\cos^2 \theta\]
anonymous
  • anonymous
Nw Form a Quadratic Equation...
mathslover
  • mathslover
I got \(\theta = \cos^{-1} \frac{1}{2}\)

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mathslover
  • mathslover
theta = 60 degrees.
mathslover
  • mathslover
but general solution???
terenzreignz
  • terenzreignz
Just add 2k(360 degrees) where k is any integer, and there's your general solution :D
anonymous
  • anonymous
\[\cos \theta =\frac{ 1 }{ 2 }\] \[Cosx=Cosy\] then \[x=(2n \Pi)\pm y\]
terenzreignz
  • terenzreignz
wait, no, k(360 degrees) not 2k sorry
calculusfunctions
  • calculusfunctions
May I explain?
mathslover
  • mathslover
Yea @calculusfunctions sir, please!
anonymous
  • anonymous
\[x= (2n \Pi) \pm \frac{ \Pi }{ 3 }\]
mathslover
  • mathslover
I had got this : cos theta = 1/2 , please explain after this.
anonymous
  • anonymous
There Will be Two value For Cos x @mathslover
mathslover
  • mathslover
Yes but the negative one will NOT BE ACCEPTED..
mathslover
  • mathslover
as cos x can NOT BE NEGATIVE @Yahoo!
anonymous
  • anonymous
Why min value of cos x = -1 ....
terenzreignz
  • terenzreignz
Must x be in degrees?
terenzreignz
  • terenzreignz
theta rather
anonymous
  • anonymous
\[\cos 180 = -1\]
calculusfunctions
  • calculusfunctions
\[\sin ^{2}\theta -2\cos \theta +\frac{ 1 }{ 4 }=0\]First multiply the equation by the least common denominator.
mathslover
  • mathslover
Wait, there will be two values for cos theta , the other one is -5/2 but -5/2 , but since the minimum value of cos theta is -1 and hence it can not be -5/2 is this what we must think abt @Yahoo! ? (sorry for my above explanation as : cos theta can never be negative as I thought that we are talking about cos(-theta) = cos theta)
mathslover
  • mathslover
cos(theta) \(\ne\) -5/2
anonymous
  • anonymous
Yup.....nw u r Correct
calculusfunctions
  • calculusfunctions
@mathslover Let me know when you're interested in learning a more efficient method.
mathslover
  • mathslover
@calculusfunctions sir, thanks a lot , I think I got it now.
mathslover
  • mathslover
Sorry but do you have any easier or another way?
calculusfunctions
  • calculusfunctions
Yes if you're interested.

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