# Find the points on the curve y=cosx/(2+sinx) at which the tangent is horizontal?

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For the tangent line to be horizontal, it's slope is zero. We also know that the tangent line to a curve, at a given point, is the derivative of the function, at that point.

We'll determine the derivative of the function using the quotient rule:

(u/v)' = (u'*v - u*v')/v^2

Let u = cos x => u' = -sin x

Let v = 2 + sin x => v' = cos x

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

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

We'll apply Pythagorean identity:

(sin x)^2 + (cos x)^2 = 1

[cos x/(2 + sin x)]' = (-2sin x - 1)/(2 + sin x)^2

But the derivative of the function must be zero.

(-2sin x - 1)/(2 + sin x)^2 = 0

Since the denominator cannot be zero, we'll cancel the numerator:

-2sin x - 1 = 0

-2sin x = 1

sin x = -1/2

x = `(-1)^(k)` arcsin (-1/2) + k`pi`

x = `(-1)^(k+1)` *(`pi` /6) + k`pi`

**Therefore, the values of x for the tangent line to the given curve is horizontal, belong to the set {`(-1)^(k+1)`*(`pi` /6) + k`pi` / k `in` Z }.**