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Math Help - An Equation

  1. #1
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    An Equation

    I had the following equation to do:

    7 sinx + 4 cosx = 8.

    My solutions (to 3 decimal places) were 0.927 and 1.106.

    Can someone verify these please? If they are not correct, could someone show me how to do it?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ty2391 View Post
    I had the following equation to do:

    7 sinx + 4 cosx = 8.

    My solutions (to 3 decimal places) were 0.927 and 1.106.

    Can someone verify these please? If they are not correct, could someone show me how to do it?
    i got the first answer, but not the second. what did you do? what range are we solving for by the way?
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  3. #3
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    Quote Originally Posted by ty2391 View Post
    I had the following equation to do:

    7 sinx + 4 cosx = 8.

    My solutions (to 3 decimal places) were 0.927 and 1.106.

    Can someone verify these please? If they are not correct, could someone show me how to do it?
    The second should be more like 1.176

    RonL
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  4. #4
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    Yup I got 1.176, just a typo.

    The way I did it was representing 7 sinx + 4 cosx as a single sine function.
    Once I did that, I just equated that to 8, and solved for x.
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  5. #5
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    7\sin x + 4\cos x = 8

    \sqrt{4^2+7^2} \left( \frac{7}{\sqrt{4^2+7^2}}\sin x + \frac{4}{\sqrt{4^2+7^2}}\cos x \right) = 8

    Let \theta = \tan^{-1} \frac{4}{7} then,

    \sqrt{4^2+7^2} (\sin x \cos \theta + \cos x \sin \theta) = 8
    Thus,
    \sin (x + \theta) = \frac{8}{\sqrt{55}}

    Now you can solve for x.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ty2391 View Post
    Yup I got 1.176, just a typo.

    The way I did it was representing 7 sinx + 4 cosx as a single sine function.
    Once I did that, I just equated that to 8, and solved for x.
    hmm. what i did was:

    7 \sin x + 4 \cos x = 8

    \Rightarrow 7 \sin x = 8 - 4 \cos x

    \Rightarrow 49 \sin^2 x = ( 8 - 4 \cos x)^2
    .
    .
    .


    i think by now you see where i am going with this. simplifying what i have above leads to a quadratic equation in cosine, and dealing with quadratics is easy

    Quote Originally Posted by ThePerfectHacker View Post
    7\sin x + 4\cos x = 8

    \sqrt{4^2+7^2} \left( \frac{7}{\sqrt{4^2+7^2}}\sin x + \frac{4}{\sqrt{4^2+7^2}}\cos x \right) = 8

    Let \theta = \tan^{-1} \frac{4}{7} then,

    \sqrt{4^2+7^2} (\sin x \cos \theta + \cos x \sin \theta) = 8
    Thus,
    \sin (x + \theta) = \frac{8}{\sqrt{55}}

    Now you can solve for x.
    interesting solution
    Last edited by ThePerfectHacker; October 31st 2007 at 07:37 PM.
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