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

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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