Ashna
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Please can you help me to find the general solution, in degrees, of the equation,
2cos^2 x=cosx?
Please include full working out and explanations as I am having difficulty with this question. Thank you very much for your kind help and assistance
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PhoenixTear
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Not sure if I've done this right; I'd be interested to see what other people got.

But I did it like this:

2(cos x)(cos x)= cos x
Divide both sides by cos x.
2cosx=1
cosx=1/2
x=pi/3
General solution is n2pi +/- pi/3
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Ashna
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The answer at the back of the book says:

x=(360n +/- 60) and x = (360n +/-90)


Not sure how the answer was obtained.

I need help/ advice on understanding the method
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sbeth
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What question is this in the book?
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Felix Felicis
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(Original post by Ashna)
Please can you help me to find the general solution, in degrees, of the equation,
2cos^2 x=cosx?
Please include full working out and explanations as I am having difficulty with this question. Thank you very much for your kind help and assistance
(Original post by PhoenixTear)
Not sure if I've done this right; I'd be interested to see what other people got.

But I did it like this:

2(cos x)(cos x)= cos x
Divide both sides by cos x.
2cosx=1
cosx=1/2
x=pi/3
General solution is n2pi +/- pi/3
One does not simply cancel cos x from both sides and lose solutions

2cos^{2}x - cosx = 0

\Rightarrow cosx(2cosx-1) = 0

work from there OP
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Ashna
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Exercise 5B Question 4 page 57
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sbeth
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I've done it like the person above^^ but just do it in degrees instead and that will get one of the answers.
So if you:
2(cosx)(cosx)=cosx
divide through by cosx:
2cosx=1
cosx=1/2
x=cos^-1(1/20)
make sure calculator is in degrees, that will give you x=60,
then you need to consider the cos graph, it is symmetrical and so you can just say it is plus or minus 60 as it needs to be in the range of -180 to 180 degress
therefore giving 360n+-60
Not sure how they got the second answer, although throughout this course i have ften found incorrect answers in this book.
Hope that makes sense!!
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Felix Felicis
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(Original post by sbeth)
I've done it like the person above^^ but just do it in degrees instead and that will get one of the answers.
So if you:
2(cosx)(cosx)=cosx
divide through by cosx:
2cosx=1
cosx=1/2
x=cos^-1(1/20)
make sure calculator is in degrees, that will give you x=60,
then you need to consider the cos graph, it is symmetrical and so you can just say it is plus or minus 60 as it needs to be in the range of -180 to 180 degress
therefore giving 360n+-60
Not sure how they got the second answer, although throughout this course i have ften found incorrect answers in this book.
Hope that makes sense!!
(Original post by Ashna)
Exercise 5B Question 4 page 57
The second answer is correct - you can't cancel cosx from both sides - you lose solutions!

x^{2} = 2x Cancel 'x' from both sides => x = 2

So x = 2 is the only solution?
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sbeth
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(Original post by Felix Felicis)
One does not simply cancel cos x from both sides and lose solutions

2cos^{2}x - cosx = 0

\Rightarrow cosx(2cosx-1) = 0

work from there OP
Just noticed this message and this will give the other answer as when c0x=0, x=90 and again cos graph symmetrical and so +/- 90 degrees
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Felix Felicis
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(Original post by sbeth)
Just noticed this message and this will give the other answer as when c0x=0, x=90 and again cos graph symmetrical and so +/- 90 degrees
posting full solutions isn't going to help OP learn, ya know...
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Ashna
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thanks very much sbeth
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Ashna
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one other thing, how is the interval determined whether it's 360 or 180?
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sbeth
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with cos and sin its always between 180 and -180 and with tan its between 90 and -90! have you got your exam today? good luck
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