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Math Help - 5 trigo problem [general solution]

  1. #1
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    5 trigo problem [general solution]

    p213 ex10a
    sorry for my laziness and bad hand writing. thanks
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  2. #2
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    sin(4x)+sin(2x)=0

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

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

    sin(2x)=0
    2x=0,\pi,2\pi, ....
    x=0,\frac{\pi}{2},\pi, .... \rightarrow \frac{n\pi}{2}

    (2cos(2x)+1)=0
    cos(2x)=\frac{-1}{2}
    2x=\frac{2\pi}{3},\frac{4\pi}{3},\frac{8\pi}{3},\f  rac{10\pi}{3}, ....
    x=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\fra  c{5\pi}{3}, ....
    x=\frac{n\pi}{3}

    My answer agrees 50% with your text's...?
    Last edited by colby2152; February 4th 2008 at 05:27 AM.
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  3. #3
    MHF Contributor red_dog's Avatar
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    18. \displaystyle\frac{(2n+1)\pi}{3}-\frac{5\pi}{36}=\frac{2n\pi}{3}+\frac{\pi}{3}-\frac{5\pi}{36}=\frac{2n\pi}{3}+\frac{7\pi}{36}.
    So, your answer is correct.
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  4. #4
    MHF Contributor red_dog's Avatar
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    19. (1+\sqrt{3})\sin\theta-(1-\sqrt{3})\cos\theta=2\sqrt{2}
    Multiply both members by \displaystyle\frac{\sqrt{2}}{4}.

    \displaystyle\frac{\sqrt{2}+\sqrt{6}}{4}\sin\theta-\frac{\sqrt{2}-\sqrt{6}}{4}\cos\theta=1

    \displaystyle\sin\frac{7\pi}{12}\sin\theta-\cos\frac{7\pi}{12}\cos\theta=-1

    \displaystyle\cos\left(\theta+\frac{7\pi}{12}\righ  t)=1

    \displaystyle\theta+\frac{7\pi}{12}=(2n+1)\pi

    \displaystyle\theta=(2n+1)\pi-\frac{7\pi}{12}=2n\pi+\frac{5\pi}{12}
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  5. #5
    MHF Contributor red_dog's Avatar
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    27. \displaystyle\sin x\cos x\cos 2x=\frac{1}{8}\Leftrightarrow8\sin x\cos x\cos 2x=1\Leftrightarrow 4\sin 2x\cos 2x=1\Leftrightarrow 2\sin 4x=1

    Then \displaystyle\sin 4x=\frac{1}{2}\Rightarrow 4x=(-1)^n\frac{\pi}{6}+n\pi\Rightarrow x=(-1)^n\frac{\pi}{24}+\frac{n\pi}{4}
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