calculus
posted by CMM .
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!

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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