Finding horizontal tangents on an interval

noelwolfe
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Homework Statement



Determine all x in [-pi/2, pi/2] at which the graph has horizontal tangents.

Homework Equations



1.) f'(x)= 9cos(x)-2sin(x)
2.) f'(x)= -5csc(x) (5cot(x)-csc(x))

The Attempt at a Solution



1.) 9cos(x)=2sin(x)
9/2=sin(x)/cos(x)
9/2=tan(x)
and then I'm not sure what to do with 9/2?

2.) 5cot(x)=csc(x)
5= csc(x)/cot(x)
5=1/cos(x)
cos(x)=1/5
same problem here... what do I do with 1/5?
 
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noelwolfe said:
9/2=tan(x)
and then I'm not sure what to do with 9/2?

You are looking for the values of x such that tan(x) = 9/2. Remember inverse trigonometric functions (i.e. sine / arcsine, cos / arccos, tan / arctan, etc.)?
 
I'm sorry, I still don't know--I'm not aware of a "simple" solution to tan(x)=9/2. How would I go about finding the solution?
 
Look up the arctan function (also called inverse tangent) and its uses.
 
Got it. Thanks!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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