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Math Help - general solution to trig problem

  1. #1
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    general solution to trig problem

    Hi,

    I'm having a lot of trouble answering this question it must be said, I've tried a number of approaches, mostly using the double angle formulae on sin4x and cos4x and then messing around with it to get it to look like something I can solve, but to no avail. Any help would be much appreciated!

    Find the general solution to

    \sin 2x + \sin 4x = \cos 2x + \cos 4x

    Regards,

    Stonehambey
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  2. #2
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    Quote Originally Posted by Stonehambey View Post
    Hi,

    I'm having a lot of trouble answering this question it must be said, I've tried a number of approaches, mostly using the double angle formulae on sin4x and cos4x and then messing around with it to get it to look like something I can solve, but to no avail. Any help would be much appreciated!

    Find the general solution to

    \sin 2x + \sin 4x = \cos 2x + \cos 4x

    Regards,

    Stonehambey

    Use the sum to product identities

    List of trigonometric identities - Wikipedia, the free encyclopedia

    2\sin\left(\frac{2x+4x}{2} \right)\cos\left(\frac{2x-4x}{2} \right)=2\cos\left(\frac{2x+4x}{2} \right)\cos\left(\frac{2x-4x}{2} \right)

    2\sin(3x)\cos(-2x)=2\cos(3x)\cos(-2x)

    2\sin(3x)\cos(2x)=2\cos(3x)\cos(2x)

    2\sin(3x)\cos(2x)-2\cos(3x)\cos(2x)=0

    2\sin(3x)\cos(2x)-2\cos(3x)\cos(2x)=0

    2\cos(2x)[\sin(3x)-\cos(3x)]=0

    You should be able to finish from here
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  3. #3
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    sum to product! Now it's obvious :P

    Thanks!
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  4. #4
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    I'm assuming you are looking for all solutions on interval [0,2pi)

    I include thr graph just to show we can expect 8 solutions

    I've included the solutions and method in the attachment .

    But the basic method i used was to rearrange as

    sin(2x)-cos(2x) = cos(4x)-sin(4x)

    Then I squared both sides and got 2 fairly simple equations

    however there are 16 possibilities of which 8 are solutions--this is the tedious part.

    If someone has an easier way I'd like to know it.

    should have waited
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  5. #5
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    hmm, I tried the way theEmptySet suggested and got a little stuck again

    I applied the formulas and got to the same bit that you did. When I solved for

    2\cos(2x)=0

    I get x=\frac{\pi}{4} + n\pi, which doesn't work when I sub back into the original equation, so where did I go wrong?
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  6. #6
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    Quote Originally Posted by Stonehambey View Post
    hmm, I tried the way theEmptySet suggested and got a little stuck again

    I applied the formulas and got to the same bit that you did. When I solved for

    2\cos(2x)=0

    I get x=\frac{\pi}{2} + n\pi, which doesn't work when I sub back into the original equation, so where did I go wrong?
    You are really close you should have gotten

    2x = \frac{\pi}{2}+n\pi \iff x =\frac{\pi}{4}+n\frac{\pi}{2}
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  7. #7
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    Quote Originally Posted by TheEmptySet View Post
    You are really close you should have gotten

    2x = \frac{\pi}{2}+n\pi \iff x =\frac{\pi}{4}+n\frac{\pi}{2}
    eek, sorry that was a typo on my part What I meant to type was

    \cos(2x) = 0

    2x = \pm \frac{\pi}{2} + 2\pi n

    \implies x = \pm \frac{\pi}{4} + n \pi

    If we let n = 0 and sub back into the original equation we have

    \sin\left(\frac{\pi}{2}\right) + \sin(\pi) = \cos\left(\frac{\pi}{2}\right) + \cos(\pi)

    1 = -1

    which is obviously wrong but I can't see where the error is

    EDIT: I got Calculus26 method to work, very nice idea to square both sides I didn't think of that. Now I'm wondering why I can't get it to work using the sum to product way =P
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  8. #8
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    Quote Originally Posted by Stonehambey View Post
    eek, sorry that was a typo on my part What I meant to type was

    \cos(2x) = 0

    2x = \pm \frac{\pi}{2} + 2\pi n

    \implies x = \pm \frac{\pi}{4} + n \pi

    If we let n = 0 and sub back into the original equation we have

    \sin\left(\frac{\pi}{2}\right) + \sin(\pi) = \cos\left(\frac{\pi}{2}\right) + \cos(\pi)

    1 = -1

    which is obviously wrong but I can't see where the error is

    EDIT: I got Calculus26 method to work, very nice idea to square both sides I didn't think of that. Now I'm wondering why I can't get it to work using the sum to product way =P
    Cosine has a zero every Pi units 2pi is too much see my above post.
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  9. #9
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    Quote Originally Posted by TheEmptySet View Post
    Cosine has a zero every Pi units 2pi is too much see my above post.
    You are right it only works for odd n.
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    Quote Originally Posted by TheEmptySet View Post
    Cosine has a zero every Pi units 2pi is too much see my above post.
    yeah, I accounted for that when I gave the general solution as

    2x = \pm\frac{\pi}{2} + 2n\pi

    starting at \frac{\pi}{2} and going up by 2\pi each time gives \frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}, ...

    Starting at -\frac{\pi}{2} and going up by 2\pi each time gives us \frac{3\pi}{2}, \frac{7\pi}{2}, \frac{11\pi}{2}, ...

    So all the zeros are covered by these two sets of solutions

    (obviously n could be negative and we get zeros going the other way as well)

    But how do I know which ones will hold true for the original equation (when I divide them by 2 to get x on its own)?
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  11. #11
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    Quote Originally Posted by TheEmptySet View Post
    You are really close you should have gotten

    2x = \frac{\pi}{2}+n\pi \iff x =\frac{\pi}{4}+n\frac{\pi}{2}
    So can any one explain why this only works for odd n? This problems is driving me up the wall!!

    Regards,

    Stonehambey
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