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calculus

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

I am not sure, but would I find the derivative first:

y'= [(2 + sin x)(-sin x) - (cos x)(cos x)]/(2 + sin x)^2

But then I don't know what to do or if that is even correct???

Would I simplify: (-sin x)*(sin x + 2) - (cos x)^2*(sin x + 2)^2

Then I would plug points in (x,y), but I am not given points.

Please Help!

  • calculus - ,

    your derivative is correct so for a tangent to be horizontal, its slope would have to be zero, so

    [(2 + sin x)(-sin x) - (cos x)(cos x)]/(2 + sin x)^2 = 0
    [2 + sin x)(-sin x) - (cos x)(cos x) = 0
    -2sinx - sin^2x - cos^2x = 0
    -2sinx -(sin^2x + cos^2x = 0
    -2sinx - 1 = 0
    sinx = - 1/2
    so x is in quadrants III or IV
    x = 210° or x = 330°
    x = 7π/6 or x = 11π/6

    sub those back in the original to get the corresponding y values, (you do it)

    points are (7π/6, ....) and (11π/6, ....)

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