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Math Help - Trigonometric General Solutions

  1. #1
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    Trigonometric General Solutions

    I guess Trigonometry question go in the Trigonometry section. That makes sense.

    I was having trouble with this question (See attachment) as I'm not sure how to change it into a sum or something similar to find the general solution. Iím not sure if I need to use identities or a product to sum formula or something similar. If someone could work me through it, it would be greatly appreciated.
    Attached Thumbnails Attached Thumbnails Trigonometric General Solutions-question-6.jpg  
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  2. #2
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    Try double angle formula
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  3. #3
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    I thought of using the double angle formula on sin4x but I'm not sure how to use this to find the general solutions. If you could show me what you mean it would help a lot.
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  4. #4
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    Please show your work first ^^

    EDIT : That's a good start to use double angle formula on sin 4x
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  5. #5
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    Using the double angle formula sin4x-cosx=0,

    =2sin2xcos2x-cosx=0

    This is all I could manage to do. I'm not sure if I should use the double angle formula on cosx either?
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  6. #6
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    Use double angle formula again on sin 2x first, then factorise the equation
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  7. #7
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    songoku,

    I'm not sure how to use the double angle formula a second time and then factorise it. It would be helpful if you could show me how to do that and then showhow to factorise it.

    Thanks.
    Last edited by Daddy_Long_Legs; July 23rd 2009 at 12:14 AM. Reason: spelling mistake
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  8. #8
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    Quote Originally Posted by Daddy_Long_Legs View Post
    I guess Trigonometry question go in the Trigonometry section. That makes sense.

    I was having trouble with this question (See attachment) as I'm not sure how to change it into a sum or something similar to find the general solution. I’m not sure if I need to use identities or a product to sum formula or something similar. If someone could work me through it, it would be greatly appreciated.
    A simple approach:

    \sin (4x) = \cos (x)

    \Rightarrow \sin (4x) = \sin \left( \frac{\pi}{2} - x \right)

    using the complementary angle formula.

    Case 1: 4x = \frac{\pi}{2} - x + 2n \pi

    Case 2: 4x = \pi - \left( \frac{\pi}{2} - x\right) + 2n \pi

    where n is an integer. Solve for x in each case.
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  9. #9
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    wow, really nice work mr_fantastic

    btw, I mean like this :
    2sin2xcos2x-cosx = 0
    4 sin x cos x cos 2x - cos x = 0
    (cos x) (4 sin x cos 2x - 1 ) = 0
    cos x = 0 or 4 sin x cos 2x - 1 = 0

    Use the double angle formula again on cos 2x and state the equation in term of sin x

    but mr_fantastic has given a really elegant way
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