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Trigonometry 7 Online
mathslover (mathslover):

find the general solution for : \(\sin^2 \theta -2 \cos \theta + \frac{1}{4} = 0 \)

OpenStudy (anonymous):

\[Sin^2\theta = 1-\cos^2 \theta\]

OpenStudy (anonymous):

Nw Form a Quadratic Equation...

mathslover (mathslover):

I got \(\theta = \cos^{-1} \frac{1}{2}\)

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

OpenStudy (anonymous):

\[\cos \theta =\frac{ 1 }{ 2 }\] \[Cosx=Cosy\] then \[x=(2n \Pi)\pm y\]

terenzreignz (terenzreignz):

wait, no, k(360 degrees) not 2k sorry

OpenStudy (calculusfunctions):

May I explain?

mathslover (mathslover):

Yea @calculusfunctions sir, please!

OpenStudy (anonymous):

\[x= (2n \Pi) \pm \frac{ \Pi }{ 3 }\]

mathslover (mathslover):

I had got this : cos theta = 1/2 , please explain after this.

OpenStudy (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!

OpenStudy (anonymous):

Why min value of cos x = -1 ....

terenzreignz (terenzreignz):

Must x be in degrees?

terenzreignz (terenzreignz):

theta rather

OpenStudy (anonymous):

\[\cos 180 = -1\]

OpenStudy (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

OpenStudy (anonymous):

Yup.....nw u r Correct

OpenStudy (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?

OpenStudy (calculusfunctions):

Yes if you're interested.

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