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Math Help - Need help on derivatives of trignometric function

  1. #1
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    Need help on derivatives of trignometric function

    Find all points on the curve below at which the tangent line is horizontal. y = (cos x) / (2 + sin x)


    can somebody help me to solve this question?



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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by btwest69 View Post
    Find all points on the curve below at which the tangent line is horizontal. y = (cos x) / (2 + sin x)


    can somebody help me to solve this question?



    Hint: the tangent line is horizontal if it has a slope of zero, that is, the derivative is zero. so find the derivative, set it equal to zero and solve for the x's
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    Quote Originally Posted by Jhevon View Post
    Hint: the tangent line is horizontal if it has a slope of zero, that is, the derivative is zero. so find the derivative, set it equal to zero and solve for the x's

    -2sinx -1 / (2+sinx)^2
    I am not sure of my answer for derivative
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by btwest69 View Post
    -2sinx -1 / (2+sinx)^2
    I am not sure of my answer for derivative
    yes it is, if by that you mean \frac {-2 \sin x - 1}{(2 + \sin x )^2}

    now set that to zero, and solve for x
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    Quote Originally Posted by Jhevon View Post
    yes it is, if by that you mean \frac {-2 \sin x - 1}{(2 + \sin x )^2}

    now set that to zero, and solve for x

    i am stuck with this step
    can you show me how to make it to 0
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by btwest69 View Post
    i am stuck with this step
    can you show me how to make it to 0
    a fraction is zero when its numerator is zero (provided its denominator is not zero at the same time). just set the numerator equal to zero and solve for x
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    Quote Originally Posted by Jhevon View Post
    a fraction is zero when its numerator is zero (provided its denominator is not zero at the same time). just set the numerator equal to zero and solve for x

    -2 \sin x - 1}{(2 + \sin x )^2=0

    my answer for x are 2 and 1/2


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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by btwest69 View Post

    -2 \sin x - 1}{(2 + \sin x )^2=0

    my answer for x are 2 and 1/2


    what?

    -2 \sin 2 - 1 \ne 0

    and -2 \sin \frac 12 - 1 \ne 0

    you have a trig equation here. the x's will be angles, and you will have infinitely many answers, not just 2

    as i said, you need to solve -2 \sin x - 1 = 0 \implies \sin x = - \frac 12

    what x's does that happen for?


    EDIT: I'm going to bed, so i won't be able to answer you if you respond, so i'll just give you the answers.

    x = \frac {7 \pi}6 + 2n \pi for n \in \mathbb{Z} or x = \frac {11 \pi}6 + 2k \pi for k \in \mathbb{Z}
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