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    So I rememeber in C2 that we couldn't divide by the trig functions cosx, sinx, tanx as that would lose solutions (because if you divide by cosx, your discarding the possible solution cosx = 0 right?)

    So how come here we're allowed to divide by cosx?

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    It isn't a factor on both sides, so you're not losing it. (It basically becomes part of tan.)
    If it was cosxsinx=cosxcosx then you wouldn't be able to divide.
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    (Original post by MathMoFarah)
    It isn't a factor on both sides, so you're not losing it. (It basically becomes part of tan.)
    If it was cosxsinx=cosxcosx then you wouldn't be able to divide.
    So is it only this question where we are allowd to divide by cosine?

    Why does this method below give me extra solutions which don't work?

    cos2\theta - sin2\theta = 0

cos2\theta - cos2\theta tan2\theta = 0

cos2\theta( 1 - tan2\theta) = 0





cos2\theta= 0 , tan2\theta= 1

2\theta = \frac{\pi}{2}, \frac{3\pi}{2} , 2\theta = \frac{\pi}{4} , \frac{5\pi}{4}



\theta = \frac{\pi}{4} , \frac{3\pi}{4}, \frac{\pi}{8} , \frac{5\pi}{8}
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    (Original post by GarlicBread01)
    So is it only this question? Why does this method give me extra solutions which dont work?

    cos2\theta= 0 , tan2\theta= 1
    If you solve these equations, you'll find that they both give you the same set of solutions.
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    (Original post by Pangol)
    If you solve these equations, you'll find that they both give you the same set of solutions.
    But it doesn't, the cosine solution produces answers which when you plug back into the original equation doesn't work out
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    (Original post by GarlicBread01)
    So I rememeber in C2 that we couldn't divide by the trig functions cosx, sinx, tanx as that would lose solutions (because if you divide by cosx, your discarding the possible solution cosx = 0 right?)

    So how come here we're allowed to divide by cosx?
    If you divide an equation by a variable, you have to be sure that the variable cannot be equal to 0 in the equation (you can't divide by 0), otherwise you'll lose correct solutions.

    \displaystyle cos 2 \theta = \sin 2 \theta

    Think about why cos 2 \theta \neq 0 in this equation. Thus it is safe to divide by \cos 2 \theta.
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    (Original post by notnek)
    If you divide an equation by a variable, you have to be sure that the variable cannot be equal to 0 in the equation (you can't divide by 0), otherwise you'll lose correct solutions.

    \displaystyle cos 2 \theta = \sin 2 \theta

    Think about why cos 2 \theta \neq 0 in this equation. Thus it is safe to divide by \cos 2 \theta.
    I see, if cos is 0 sin is not zero so they can't be equal. Is this a special case then or are there other common cases where you know you can divide without losing solutions? How would I be able to see this in an exam?

    Also, what I don't understand is how did my method produce extra solutions which were wrong if every step is mathematically correct?
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    (Original post by GarlicBread01)
    So is it only this question where we are allowd to divide by cosine?

    Why does this method below give me extra solutions which don't work?

    cos2\theta - sin2\theta = 0

cos2\theta - cos2\theta tan2\theta = 0

cos2\theta( 1 - tan2\theta) = 0





cos2\theta= 0 , tan2\theta= 1

2\theta = \frac{\pi}{2}, \frac{3\pi}{2} , 2\theta = \frac{\pi}{4} , \frac{5\pi}{4}



\theta = \frac{\pi}{4} , \frac{3\pi}{4}, \frac{\pi}{8} , \frac{5\pi}{8}
    Your second line : \sin 2 \theta = \cos 2 \theta \tan 2\theta

    This has come from using \displaystyle \tan x =\frac{\sin 2\theta}{\cos 2\theta}, which is only true if we assume \cos 2\theta \neq 0 (can't divide by 0).

    So \cos 2\theta = 0 produces invalid solutions.


    A simpler example:

    x = 1

    \Rightarrow \displaystyle x = \frac{1}{x} \times x

    \Rightarrow x^2 = x

    \Rightarrow x = 0, \ x=1
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    (Original post by GarlicBread01)
    I see, if cos is 0 sin is not zero so they can't be equal. Is this a special case then or are there other common cases where you know you can divide without losing solutions? How would I be able to see this in an exam?
    Most of the time you'll be fine if you follow the rule : never divide by a trig function in an equation.

    Also, it's useful to think about assumptions when using identities

    e.g. \displaystyle \tan x = \frac{\sin x}{\cos x}

    This is not true when \cos x = 0. So you have to assume \cos x \neq 0 when using this identity.
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    (Original post by GarlicBread01)
    So is it only this question where we are allowd to divide by cosine?

    Why does this method below give me extra solutions which don't work?

    cos2\theta - sin2\theta = 0

cos2\theta - cos2\theta tan2\theta = 0

cos2\theta( 1 - tan2\theta) = 0





cos2\theta= 0 , tan2\theta= 1

2\theta = \frac{\pi}{2}, \frac{3\pi}{2} , 2\theta = \frac{\pi}{4} , \frac{5\pi}{4}



\theta = \frac{\pi}{4} , \frac{3\pi}{4}, \frac{\pi}{8} , \frac{5\pi}{8}
    I posted something about this problem some time ago:

    http://www.thestudentroom.co.uk/show...02&postcount=4
 
 
 
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