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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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??

Mathematics
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how did you get pi/3 and 5pi/3?
I basically equaled it to zero and solved from there
equalled to 0? \(\bf cos(t)=2sin^2(t)\implies cos(t)-2sin^2(t)=0?\)

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Yepp thats what I did except its 2sin squared t - cos t=0
hmm I don't see how pi/3 and 5pi/3 comes out of that though
@jdoe0001 cos theta of 1/2 equals to pi/3 or cos =-1 which is 5pi/3
hmmm well, let us start off with the pythagorean identity of \(\bf sin^2(\theta)+cos^2(\theta)=1\ so \(\bf cos(t)=2sin^2(t)\implies cos(t)-2sin^2(t)=0 \\ \quad \\ \textit{thus we could say that }sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)={\color{brown}{ 1-cos^2(\theta) }} \\ \quad \\ thus\implies cos(t)-2sin^2(t)=0\implies cos(t)-2[{\color{brown}{ 1-cos^2(t) }}]=0 \\ \quad \\ cos(t)-2+2cos^2(t)=0\implies 2cos^2(t)+cos(t)-2=0\) notice, is a quadratic equation it doesn't factor out nicely, thus you'd need to use the quadratic formula
We can't use quadtric formula tho :///
can't? if it's a quadratic one.. why not?
@jdoe0001 cuz my professor only said to use trig functions which is where I am running into problems
well... 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
@jdoe0001 Yeah it is I forgot to put the one!
well.. you may want to repost with the proper expression then, so we can see it

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