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November 28, 2014

November 28, 2014

Posted by **CMM** on Sunday, October 24, 2010 at 3:51pm.

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 -
**Reiny**, Sunday, October 24, 2010 at 4:29pmyour 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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