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anonymous
 one year ago
If 0 is less than t less than 2 pi solve cos(t)+1=2sin^2(t)?
I got three different answers, so I got pi/3, pi, and 5pi/3
Is this right at all??
anonymous
 one year ago
If 0 is less than t less than 2 pi solve cos(t)+1=2sin^2(t)? I got three different answers, so I got pi/3, pi, and 5pi/3 Is this right at all??

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how did you get pi/3 and 5pi/3?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I basically equaled it to zero and solved from there

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0equalled to 0? \(\bf cos(t)=2sin^2(t)\implies cos(t)2sin^2(t)=0?\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yepp thats what I did except its 2sin squared t  cos t=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm I don't see how pi/3 and 5pi/3 comes out of that though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@jdoe0001 cos theta of 1/2 equals to pi/3 or cos =1 which is 5pi/3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm well, let us start off with the pythagorean identity of \(\bf sin^2(\theta)+cos^2(\theta)=1\ so \(\bf cos(t)=2sin^2(t)\implies cos(t)2sin^2(t)=0 \\ \quad \\ \textit{thus we could say that }sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)={\color{brown}{ 1cos^2(\theta) }} \\ \quad \\ thus\implies cos(t)2sin^2(t)=0\implies cos(t)2[{\color{brown}{ 1cos^2(t) }}]=0 \\ \quad \\ cos(t)2+2cos^2(t)=0\implies 2cos^2(t)+cos(t)2=0\) notice, is a quadratic equation it doesn't factor out nicely, thus you'd need to use the quadratic formula

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We can't use quadtric formula tho :///

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can't? if it's a quadratic one.. why not?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@jdoe0001 cuz my professor only said to use trig functions which is where I am running into problems

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well... for this quadratic... it won't give you any nice integers so... I'd use the quadratic... unless, your expression is different from what you posted

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@jdoe0001 Yeah it is I forgot to put the one!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well.. you may want to repost with the proper expression then, so we can see it
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