Here's the question you clicked on:
trusev1
Find all points (i.e. all x-values) between 0 and 2pi where the line tangent to the graph of y =sinx/2+cosx is horizontal
Have you considered the 1st Derivative?
1st derivative would be -(cot x) (cosx/-sinx) how would I get the points from that?
If that is the correct 1st derivative, which I didn't check, you determine where it is zero.
well the derivative of sinx is cosx and the derivative of any constant is always 0 and the derivative of cosx is -sinx and we know that cotx =cosx/sinx
I am not sure how I would determine where it would be zero..
horizontal tg = 0(no variation)
No. That will not do. You need the Quotient Rule. I get the 1st Derivative as \(\dfrac{2\cos(x)+1}{(2+\cos(x))^{2}}\). After that, simply concern yourself with the numerator being zero, unless the denominator happens to be zero in the same place.
This assumes your original expression was \(\dfrac{\sin(x)}{2+\cos(x)}\). Was it?
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yes tkhunny that was the original equation
Then I gave the correct 1st Derivative and you should use parentheses to clarify your meaning. Remember your Order of Operations.
rapahael can you still help?
so do I need to set the numerator =0
yes, because if the tangent line is horizontal, his angular coefficient is zero
so then I end up with cosx=-1/2
so is that my only x value , -1? ( my reasoning behind that is that cos of 60degrees=1/2) so then y x value is -1?
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pi-pi/3,pi+pi/3
@trusev1 get it?
I get that but it is asking for the x-values that are at the line tangent is horizontal. so we got the derivative and even from your picture both those points would have the same x-value.
no. my figure isnt the graph of your question,it's just the trigonometric circle