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confused about this c2 trig question

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    so the question is solve (1+cosx)^2= 1/4 for -180<x<180

    I know that what you would have to do square root both sides and go on from there. but i did something different which should give the same answer but didn't

    I squared it so i did (1+cosx)(1+cosx) to give me 1+2cosx+cos^2x=1/4

    then i took the one to the other side to give me 2cosx+cos^2x=-3/4

    then i factorised it to give me cosx(2+cosx)= -3/4

    so cosx=-3/4 and 2+cosx=-3/4

    doing x=cos^-1(-3/4) gives me 138.5903779
    and doing x=cos^-1(-11/4) gives me a maths error

    i went on further with the first one which gave me a solution but in fact the answer is +-120, i understand how that answer was obtained as you simply square root both sides and go on from there, but why was the method that i did (which should have given me the right answer) faulty?
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    You would have been correct had you taken the -3/4 back to the other side - you should then be able to see that this looks like a normal quadratic equation in cosx.

    I think your method is faulty because what you have done isn't actually factorising - you could try saying "let cosx = t" to make it easier to see the equation as a more normal looking quadratic.
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    (Original post by `God)
    so the question is solve (1+cosx)^2= 1/4 for -180<x<180

    I know that what you would have to do square root both sides and go on from there. but i did something different which should give the same answer but didn't

    I squared it so i did (1+cosx)(1+cosx) to give me 1+2cosx+cos^2x=1/4

    then i took the one to the other side to give me 2cosx+cos^2x=-3/4

    then i factorised it to give me cosx(2+cosx)= -3/4

    so cosx=-3/4 and 2+cosx=-3/4

    doing x=cos^-1(-3/4) gives me 138.5903779
    and doing x=cos^-1(-11/4) gives me a maths error

    i went on further with the first one which gave me a solution but in fact the answer is +-120, i understand how that answer was obtained as you simply square root both sides and go on from there, but why was the method that i did (which should have given me the right answer) faulty?

    Highlighted in bold - that bit isn't true! It isn't possible to deduce that Cos x = -3/4 from this equation - because any two numbers could multiply to give -3/4.
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    (Original post by PerArduaAdAstra)
    You would have been correct had you taken the -3/4 back to the other side - you should then be able to see that this looks like a normal quadratic equation in cosx.

    I think your method is faulty because what you have done isn't actually factorising - you could try saying "let cosx = t" to make it easier to see the equation as a more normal looking quadratic.
    i think the factorising is correct..
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    Fully factorizing one side of the equation only works if the other side is 0.

    This is because the step you are taking is:

    \alpha\beta\gamma=0 \Leftrightarrow \alpha=0\ \text{OR}\ \beta=0\ \text{OR}\ \gamma=0

    This is because of the Multiplicative Property of Zero (things multiplied together always equal zero iff at least one of them is zero).
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    (Original post by Funkyhouse)
    tHis is the quickest way square root both sides(1+cosx)^2=1/4 so 1 + cosx =sq rt of 1/4 which is 1/2 then minus 1 from both sides leaving u with cosx = -1/2 the do the inverse arcos-1/2 which gives u 120+-
    yes i know that i longed it out, but i don't know where i went wrong
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    (Original post by aznkid66)
    Fully factorizing one side of the equation only works if the other side is 0.

    This is because the step you are taking is:

    \alpha\beta\gamma=0 \Leftrightarrow \alpha=0\ \text{OR}\ \beta=0\ \text{OR}\ \gamma=0

    This is because of the Multiplicative Property of Zero (things multiplied together always equal zero iff at least one of them is zero).
    oh i never knew that, and that applies for everything?
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    (Original post by `God)
    oh i never knew that, and that applies for everything?
    Yup. If you don't set an equation to zero, you should have better luck simplifying rather than factorizing.
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    (Original post by aznkid66)
    Fully factorizing one side of the equation only works if the other side is 0.

    This is because the step you are taking is:

    \alpha\beta\gamma=0 \Leftrightarrow \alpha=0\ \text{OR}\ \beta=0\ \text{OR}\ \gamma=0

    This is because of the Multiplicative Property of Zero (things multiplied together always equal zero iff at least one of them is zero).
    Aaah that's really interesting i thought of that but didn't know how to explain it properly :adore:
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    (Original post by Funkyhouse)
    Aaah that's really interesting i thought of that but didn't know how to explain it properly :adore:
    To be fair, Blazy said it before me, and I think PAAA was hinting at it ^^
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    (Original post by aznkid66)
    To be fair, Blazy said it before me, and I think PAAA was hinting at it ^^
    Ah modesty, but yh I agree with you :adore:Blazy, :adore:PAAA
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    (Original post by f1mad)
    I'm not quite sure why you expanded that whole thing:

    (1+cosx)^2= 1/4

    Square root both sides.
    As stated in the OP, he was trying to be different, and the problem was that he didn't know the mistake in his method.

    There's nothing wrong with expanding, anyways.

    1+2cosx+cos^2=1/4

    3/4+2cosx+cos^2=0
    (1/2+cosx)(3/2+cosx)=0
    cosx=1/2, 3/2
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    (Original post by `God)
    so the question is solve (1+cosx)^2= 1/4 for -180<x<180

    I know that what you would have to do square root both sides and go on from there. but i did something different which should give the same answer but didn't

    I squared it so i did (1+cosx)(1+cosx) to give me 1+2cosx+cos^2x=1/4

    then i took the one to the other side to give me 2cosx+cos^2x=-3/4

    then i factorised it to give me cosx(2+cosx)= -3/4

    so cosx=-3/4 and 2+cosx=-3/4

    doing x=cos^-1(-3/4) gives me 138.5903779
    and doing x=cos^-1(-11/4) gives me a maths error

    i went on further with the first one which gave me a solution but in fact the answer is +-120, i understand how that answer was obtained as you simply square root both sides and go on from there, but why was the method that i did (which should have given me the right answer) faulty?
    Do you have a graphical calculator? I find, especially in exams, plotting the graph and finding the areas of intersection helps to see if the calculations are correct. You do still get marks for working, which is great, but double checking your answer against the calculator is great too . If you don't have one of these calculators, as expensive as they are, they're extremely helpful
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    (Original post by `God)
    oh i never knew that, and that applies for everything?
    :eek:

    If (x-2)(x+1) = 0

    (x-2) = 0 means x = 2 giving 0*3 = 0
    (x+1) = 0 means x = -1 giving -3*0 = 0

    So these values of x work

    HOWEVER

    If (x-2)(x+1) = 5

    lets try your approach

    (x-2) = 5 means x = 7 giving 5*8 = 40 ........ OOOPS

    So you cannot use the same approach

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