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AQA FP1 Trig help

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
(edited 11 years ago)
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
Reply 2
Avatar for Ashna
Ashna
OP
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
Reply 3
Avatar for sbeth
sbeth
5
What question is this in the book?
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

2cos2x−cosx=02cos^{2}x - cosx = 0

⇒cosx(2cosx−1)=0\Rightarrow cosx(2cosx-1) = 0

work from there OP
Reply 5
Avatar for Ashna
Ashna
OP
Exercise 5B Question 4 page 57
Reply 6
Avatar for sbeth
sbeth
5
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 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 cosxcosx from both sides - you lose solutions!

x2=2xx^{2} = 2x Cancel 'x' from both sides => x=2x = 2

So x = 2 is the only solution?
Reply 8
Avatar for sbeth
sbeth
5
Original post by Felix Felicis
One does not simply cancel cos x from both sides and lose solutions

2cos2x−cosx=02cos^{2}x - cosx = 0

⇒cosx(2cosx−1)=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:smile:
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:smile:

posting full solutions isn't going to help OP learn, ya know...
Reply 10
Avatar for Ashna
Ashna
OP
thanks very much sbeth
Reply 11
Avatar for Ashna
Ashna
OP
one other thing, how is the interval determined whether it's 360 or 180?
Reply 12
Avatar for sbeth
sbeth
5
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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