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Math Help - Identities and Letting Variables equal

  1. #1
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    Identities and Letting Variables equal

    I'm having trouble figuring out how to solve cos x = 2sinxcosx within 0 to 2pi

    I know that 2sinxcosx is an identity for sin2x but that doesn't really help the situation since I want to let cosx be a variable (ie. let cosx = m and then solve for roots/angles)

    I was thinking about factoring the second side of the equation but it seems like I can't get anything out of that either, I would just get

    cos x = sinx (2cosx)
    or
    cos x = cosx (2sinx)

    which doesn't really do anything Anyways if anyone can help solve it by letting either cosx or sinx being a variable it would greatly be appreciated The answer is supposedly 3pi/2, thanks for the help in advance!
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by peekay View Post
    I'm having trouble figuring out how to solve cos x = 2sinxcosx within 0 to 2pi

    I know that 2sinxcosx is an identity for sin2x but that doesn't really help the situation since I want to let cosx be a variable (ie. let cosx = m and then solve for roots/angles)

    I was thinking about factoring the second side of the equation but it seems like I can't get anything out of that either, I would just get

    cos x = sinx (2cosx)
    or
    cos x = cosx (2sinx)

    which doesn't really do anything Anyways if anyone can help solve it by letting either cosx or sinx being a variable it would greatly be appreciated The answer is supposedly 3pi/2, thanks for the help in advance!

    sin(2x) doesn't come into use here, this is a factorising question

    Method 1

    Take cos(x) from both sides:

    2sin(x)cos(x) - cos(x) = 0

    Factor out cos(x):

    cos(x)(2sin(x)-1) = 0

    Either cos(x) = 0 \rightarrow x = \frac{\pi}{2}

    Or sin(x) = \frac{1}{2} \rightarrow x = \frac{\pi}{6}

    Since cos(x) is symmetrical about \pi then \frac{\pi}{2} + \pi = \frac{3\pi}{2} is also a solution

    Also note that because 2sin(x) is symmetrical about pi/2 then it follows that there is another root at \frac{\pi}{2}+({\pi}{2} - \frac{\pi}{6}) = \frac{5\pi}{6}.

    Therefore the equation 2sin(x)cos(x) = cos(x) has roots of \frac{\pi}{6} \: , \: \frac{\pi}{2} \: , \: \frac{5\pi}{6} \: and \: \frac{3\pi}{2}

    ---------

    Method 2

    Divide through by cos(x)

    2sin(x) = 1

    sin(x) = \frac{1}{2}

    x = \frac{\pi}{6} \: , \: \frac{5\pi}{6}

    Note that method 1 is better because it yielded two extra solutions
    Last edited by e^(i*pi); June 20th 2009 at 04:20 PM.
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  3. #3
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    Thanks for both the solutions, really do appreciate it.

    I like method 2 cause you can just spot it out right away, but considering I'll be writing an exam on monday I better try and get all the solutions I can get .
    Last edited by peekay; June 20th 2009 at 04:29 PM.
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  4. #4
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    Quote Originally Posted by e^(i*pi) View Post
    sin(2x) doesn't come into use here, this is a factorising question

    Method 1

    Take cos(x) from both sides:

    2sin(x)cos(x) - cos(x) = 0

    Factor out cos(x):

    cos(x)(2sin(x)-1) = 0

    Either cos(x) = 0 \rightarrow x = \frac{\pi}{2}

    Or sin(x) = \frac{1}{2} \rightarrow x = \frac{\pi}{6}

    Since cos(x) is symmetrical about \pi then \frac{\pi}{2} + \pi = \frac{3\pi}{2} is also a solution

    Also note that because 2sin(x) is symmetrical about pi/2 then it follows that there is another root at \frac{\pi}{2}+({\pi}{2} - \frac{\pi}{6}) = \frac{5\pi}{6}.

    Therefore the equation 2sin(x)cos(x) = cos(x) has roots of \frac{\pi}{6} \: , \: \frac{\pi}{2} \: , \: \frac{5\pi}{6} \: and \: \frac{3\pi}{2}

    ---------

    Method 2

    Divide through by cos(x)

    2sin(x) = 1

    sin(x) = \frac{1}{2}

    x = \frac{\pi}{6} \: , \: \frac{5\pi}{6}

    Note that method 1 is better because it yielded two extra solutions
    Method 2 is not valid.

    You're dividing by a quantity that potentially can equal zero. So not only are you comitting an invalid operation, you're losing (as you point out yourself) potential solutions.


    Here is a valid alternative method (first read this thread: http://www.mathhelpforum.com/math-he...lution-qs.html):

    \cos (x) = \sin (2x) \Rightarrow \sin\left( \frac{\pi}{2} - x \right) = \sin (2x).

    Case 1: \frac{\pi}{2} - x = 2x + 2n \pi \Rightarrow x = \frac{\pi}{6} - \frac{2n \pi}{3} where n is an integer.


    Case 2: \frac{\pi}{2} - x = (\pi - 2x) + 2n \pi \Rightarrow x = (2n + 1) \pi - \frac{\pi}{2} where n is an integer.
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