how to find horizontal tangent

airz said:

Hi,
here's the question im trying to answer:

Find the points on the curve y= (cos x)/(2 + sin x) at which the tangent is horizontal.

I've found the derivative, which is -(sin^2 x + 2sin x + cos^2 x)/(sin^2 x + 4sin x + 4) but what do I do from there? Answers to similar questions in the back of the book are expressed with pi and I have no idea how to get pi from an expression like that.

Thanks a lot!

How is the derivative at a point is related to the slope of the tangent at that point?

If you don't know - look up in your text-book - if no text-book use google.

Next - what is the slope of a horizontal line?

airz said:

the derivative at the point is equal to the slope of the tangent line, and the slope of a horizontal line is zero. so, I'm looking for the values where the derivative equals zero. how can I simplify that expression when I set it equal to zero, though?

thanks !

That is simple algebra - for a fractional expression to be zero the numerator must be equal to zero.

From there use trigonometry ...

Show some work - and then talk about where you are stuck...

airz said:

I think it's the trigonometry part that's getting me mixed up, it's been quite a while since I took that class. Ive been able to solve other problems like this that don't involve trig functions

So, I just set -(sin^2 x + 2sin x + cos^2 x) = 0

but I'm not sure how to solve from there.

\(\displaystyle sin^2(x) \ + \ 2sin(x) + cos^2(x) \ = \ 0\)

\(\displaystyle sin^2(x) \ + cos^2(x) \ + \ 2sin(x) \ = \ 0\)

\(\displaystyle 1 \ + \ 2sin(x) \ = \ 0\)

\(\displaystyle sin(x) \ = \ - \frac{1}{2}\)

\(\displaystyle x = [- \frac{\pi}{6} + 2n \cdot \pi] \text{and} \ [\frac{\pi}{6} + (2n \ + \ 1)\cdot \pi]\)

\(\displaystyle f(x) \ = \ \frac{cos(x)}{2+sin(x)}\)

\(\displaystyle f'(x) \ = \ \frac{[2+sin(x)][-sin(x)]-[cos(x)][cos(x)]}{[2+sin(x)]^2}\)

\(\displaystyle f'(x) \ = \ \frac{-2sin(x)-1}{[2+sin(x)]^2} \ = \ 0, \ sin(x) \ = \ \frac{-1}{2}, \ x \ = \ \frac{7\pi}{6}, \ \frac{11\pi}{6}\)

\(\displaystyle f(7\pi/6) \ = \ -\sqrt3/3, \ f(11\pi/6) \ = \ \sqrt3/3, \ y \ = \ \pm \ \sqrt3/3\)

\(\displaystyle See \ graph\)

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