Isolating x Within a Tough Derivative

NicoleLee

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Jul 23, 2014
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I am working on a question to find the coordinates of the point(s) along a curve at which it is horizontal. The equation of the curve is y = (cosx)/(2+sinx), and 0 </= x </= 2pi.

I calculated the derivative to be y' = ((2+sinx)(-sinx) - (cosx)^2) / (2+sinx)^2. And to find the points at which the curve is horizontal, I set the derivative = 0, and tried to solve for x. But I cant seem to isolate x, I am wondering if there is something I am missing, or an alternate method I should be using? Any help would be appreciated.
 
I am working on a question to find the coordinates of the point(s) along a curve at which it is horizontal. The equation of the curve is y = (cosx)/(2+sinx), and 0 </= x </= 2pi.

I calculated the derivative to be y' = ((2+sinx)(-sinx) - (cosx)^2) / (2+sinx)^2. And to find the points at which the curve is horizontal, I set the derivative = 0, and tried to solve for x. But I cant seem to isolate x, I am wondering if there is something I am missing, or an alternate method I should be using? Any help would be appreciated.

\(\displaystyle \dfrac{[2+sin(x)]*[-sin(x)]-[cos^2(x)]}{[2+sin(x)]^2} \ = \ 0 \) →

\(\displaystyle \dfrac{-2*sin(x)-sin^2(x)-cos^2(x)}{[2+sin(x)]^2} \ = \ 0 \) →

\(\displaystyle \dfrac{-2*sin(x)-1}{[2+sin(x)]^2} \ = \ 0 \) →

\(\displaystyle -2*sin(x)-1 \ = \ 0 \) →

\(\displaystyle sin(x) \ = \ -\dfrac{1}{2} \) → continue......
 
thanks very much

\(\displaystyle \dfrac{[2+sin(x)]*[-sin(x)]-[cos^2(x)]}{[2+sin(x)]^2} \ = \ 0 \) →

\(\displaystyle \dfrac{-2*sin(x)-sin^2(x)-cos^2(x)}{[2+sin(x)]^2} \ = \ 0 \) →

\(\displaystyle \dfrac{-2*sin(x)-1}{[2+sin(x)]^2} \ = \ 0 \) →

\(\displaystyle -2*sin(x)-1 \ = \ 0 \) →

\(\displaystyle sin(x) \ = \ -\dfrac{1}{2} \) → continue......


Thanks very much, I had forgotten about the cos(x)^2 + sin(x)^2 = 1 rule.
 
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