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Thread: Trigonometric Equation

  1. #1
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    Trigonometric Equation

    I need to solve $\displaystyle 4\sin\theta + 5\cos\theta =6$

    $\displaystyle 4(1-\cos^2\theta)+5\cos\theta -6=0$

    $\displaystyle -4\cos^2\theta +5\cos\theta -2=0$

    $\displaystyle \cos\theta = \frac{-5\pm \sqrt{25-4(-2)(-4)}}{8}$

    This gives me complex roots. In my book they solve the equation as follows:

    $\displaystyle 4\sin\theta=6-5\cos\theta$

    $\displaystyle 16\sin^2\theta =36-60\cos\theta +25\cos^2\theta $

    $\displaystyle 16-16\cos^2\theta = 36-60\cos\theta +25\cos^2\theta $

    $\displaystyle 0=41\cos^2\theta -60\cos\theta +20$

    $\displaystyle \cos\theta = \frac{60 \pm\sqrt{3,600-4(41)(20)}}{2(41)}$

    =$\displaystyle 0.9499 or 0.5136$

    How was my approach to the problem wrong?
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  2. #2
    Senior Member pacman's Avatar
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    your identity is suspect: sin^2 A + cos^2 A = 1

    sin^2 A = 1 - cos^2 A

    see your mistake?
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  3. #3
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    , ok I see it now. Thanks
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  4. #4
    Senior Member pacman's Avatar
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    see my graph, it will help you locate the roots
    Attached Thumbnails Attached Thumbnails Trigonometric Equation-dfg.gif  
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