Posted by **holly** on Monday, February 6, 2012 at 11:21pm.

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 -
**Reiny**, Monday, February 6, 2012 at 11:39pm
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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