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Thread: Solving equations

  1. #1
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    Solving equations

    Solve the following equation giving values for 0-360deg
    cosA=cos2A+cos4A

    cos2A=cos(30-A)

    i probably missed something on the identities part.
    thanks
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  2. #2
    MHF Contributor alexmahone's Avatar
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    $\displaystyle cos A=cos 2A+cos 4A$

    $\displaystyle cos A=2cos 3A cos A$

    $\displaystyle cos A=0$ or $\displaystyle cos 3A=\frac{1}{2}$

    $\displaystyle A=90^0$ or $\displaystyle 3A=60^0$

    $\displaystyle A=90^0$ or $\displaystyle A=20^0$
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  3. #3
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    Quote Originally Posted by arze View Post
    Solve the following equation giving values for 0-360deg
    cosA=cos2A+cos4A
    Hi

    You can use the trig identity $\displaystyle \cos p + \cos q = 2 \cos \left(\frac{p+q}{2}\right) \: \cos \left(\frac{p-q}{2}\right)$
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  4. #4
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    ah! that's what i missed in the first one. thanks!
    how about the second one?
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  5. #5
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by arze View Post
    ah! that's what i missed in the first one. thanks!
    how about the second one?
    $\displaystyle cos 2A = cos (30-A)$

    $\displaystyle 2A=30^o-A$

    $\displaystyle 3A=30^o$

    $\displaystyle A=10^o$

    OR

    $\displaystyle 2A=A-30$

    $\displaystyle A=-30^0$
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  6. #6
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    Quote Originally Posted by alexmahone View Post
    $\displaystyle cos 2A = cos (30-A)$

    $\displaystyle 2A=30^o-A$

    $\displaystyle 3A=30^o$

    $\displaystyle A=10^o$

    OR

    $\displaystyle 2A=A-30$

    $\displaystyle A=-30^0$
    i'm supposed to give the answers $\displaystyle 10^o$, $\displaystyle 130^o$, $\displaystyle 250^o$, and $\displaystyle 330
    ^o$
    i can understand $\displaystyle 10^o$ and $\displaystyle 330^o$ but what about the other two?
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  7. #7
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    Quote Originally Posted by arze View Post
    i'm supposed to give the answers $\displaystyle 10^o$, $\displaystyle 130^o$, $\displaystyle 250^o$, and $\displaystyle 330
    ^o$
    i can understand $\displaystyle 10^o$ and $\displaystyle 330^o$ but what about the other two?
    Looking back at alexmahone's work:
    Quote Originally Posted by alexmahone View Post
    $\displaystyle cos 2A = cos (30-A)$

    $\displaystyle 2A=30^o-A$

    $\displaystyle 3A=30^o$
    If A is supposed to be between 0 and 360, then 3A should be between 0 and 1080 (360 times 3). So from here you also have to mention that

    $\displaystyle 3A=390^{\circ}$ and

    $\displaystyle 3A=750^{\circ}$

    because these angles are coterminal to 30 and they are between 0 and 1080. Now divide both sides of both equations by 3 and you will get

    $\displaystyle A=130^{\circ}$ and

    $\displaystyle A=250^{\circ}$.


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  8. #8
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    how could i have missed that?
    thank you very much!
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