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calculus

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Find the points on the curve below at which the tangent is horizontal. Use n as an arbitrary integer. (Enter your answers as a comma-separated list.)
y = sin x/
2 + cos x

  • calculus - ,

    did you mean
    y = sinx/(2+cosx) ??

    I will assume you did.

    dy/dx = ( (2+cosx)(cos) - sinx(-sinx) )/(2+cosx)^2
    = (2cosx + cos^2 x + sin^2 x)/(2+cosx)^2
    = (1 + 2cosx)/(2+cosx)^2
    for a horizontal tangent, dy/dx = 0
    or
    1 + 2cosx = 0
    cosx = -1/2
    x = 2π/3 or x = 4π/3 , (120° or 240° )
    the period of cosx = 2π

    so general solutions:
    x = 2π/3 + n(2π) or
    x = 4π/3 + n(2π)

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