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Math Help - Trigonometric equation

  1. #1
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    Trigonometric equation

    solve: cos(5x) + sin(x) = 0

    I have that:

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

    therefore:

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

    from here I'm stuck... any help would be much appreciated
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Trigonometric equation

    Let's look at the first factor and consider where:

    \sin(\theta)=-\cos(\theta)

    Divide through by \cos(\theta) and so obviously \cos(\theta)\ne0:

    \tan(\theta)=-1

    \theta=\frac{3\pi}{4}+k\pi=\frac{3\pi}{4}(4k+1) where k\in\mathbb{Z}.

    So, what is \theta in this case?

    Can you use a similar method for the first factor?
    Thanks from kinhew93
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  3. #3
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    Re: Trigonometric equation

    Quote Originally Posted by MarkFL2 View Post

    Divide through by \cos(\theta) and so obviously \cos(\theta)\ne0:
    Do you mean that it is ok to divide by cos because it can't equal 0? (and therefore no solutions will be lost)
    Last edited by kinhew93; January 8th 2013 at 10:08 AM.
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Trigonometric equation

    Yes, when cos(x) = 0 then sin(x) = 1 and factors in the form we have cannot be zero.
    Thanks from kinhew93
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  5. #5
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    Re: Trigonometric equation

    Quote Originally Posted by MarkFL2 View Post
    Yes, when cos(x) = 0 then sin(x) = 1 and factors in the form we have cannot be zero.
    when cos(x) + sin(x) = 0 I can see how cos(x) cannot equal 0.

    But when written as cos(x)(1 + tan(x)) = 0 cos(x) can be 0?

    Could someone please try to explain this to me :}
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  6. #6
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    Re: Trigonometric equation

    Cos 5x + sin x = 0
    OR
    cos 5x + cos ( π/2- x )=0 [cos ( π/2- x )=sin⁡x ]
    2 cos (5x+ π/2- x )/2 cos (5x-( π/2- x ) )/2=0
    [ Because cos C + cos D = 2 cos (C+D)/2 cos (C-D)/2]
    That gives
    2 cos (4x+ π/2 )/2 cos (6x- π/2 )/2=0
    Now from here you can solve the equation.
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