Tangent Line and Coordinates of Trigonometric Function

gabyoh23
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Homework Statement



There are infinitely many points on the curve y = \frac{sin x}{\sqrt{2}- cos x} at which the tangent line to this curve is horizontal. Find the x- and y-coordinates of one such point.

Homework Equations



y' = slope of the tangent line
Etc., etc.

The Attempt at a Solution


I know you have to take the derivative of the given equation, and at first, I tried using the quotient rule, but I got nowhere with that. Then I tried rationalizing the \sqrt{2}, but that didn't really get me anywhere either. I also have no idea how to find the x- and y-coordinates.

All help is greatly appreciated!
 
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Nothing needs to be done with \sqrt{2}, it is a constant. What's the derivative of a constant? (btw, \sqrt{2} is an irrational number which is why you couldn't rationalize it.)

The coordinates are the x and y-values of the function and y=f(x) so, (x,y)=(x,f(x))
 
To the OP, please show us your work in taking the derivative using the quotient rule. I did the same and got a particularly pleasing answer.
 
I got something like \frac{1-cos^2(x)}{2-cos^2(x)}.
I'm not sure if this right at all. My prowess with the quotient rule is shoddy at best.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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