AQA FP1 Trig helpWatch

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Thread starter 6 years ago
#1
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
0
6 years ago
#2
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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Thread starter 6 years ago
#3
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
0
6 years ago
#4
What question is this in the book?
0
6 years ago
#5
(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  work from there OP
0
Thread starter 6 years ago
#6
Exercise 5B Question 4 page 57
0
6 years ago
#7
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!!
0
6 years ago
#8
(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 from both sides - you lose solutions! Cancel 'x' from both sides => So x = 2 is the only solution?
0
6 years ago
#9
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 0
6 years ago
#10
(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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Thread starter 6 years ago
#11
thanks very much sbeth
0
Thread starter 6 years ago
#12
one other thing, how is the interval determined whether it's 360 or 180?
0
6 years ago
#13
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
0
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