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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.0@jdoe0001 I wrote it a little wrong!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well... your pi/3 is correct and your 5pi/3 is also correct so.... notice your range is from \(\bf cos(t)+1=2sin^2(t)\implies cos(t)+1=2[{\color{brown}{ 1cos^2(t)}}] \\ \quad \\ cos(t)+1=22cos^2(t)\implies 2cos^2(t)+cos(t)1=0 \\ \quad \\ \textit{factoring, we get}\\ \quad \\ [2cos(t)1][cos(t)+1]=0\implies \begin{cases} 2cos(t)1=0\\ 2cos(t)=1\\ cos(t)=\frac{1}{2}\\ t=cos^{1}\left( \frac{1}{2} \right) \\\hline\\ cos(t)+1=0\\ cos(t)=1\\ t=cos^{1}(1) \end{cases}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cosine is 1 at \(\pi\) thus \(\large \measuredangle t= \begin{cases} \frac{\pi }{3}\\ \pi \\ \frac{5\pi }{3} \end{cases}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@jdoe0001 But we cant use quadratic formula tho?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0why not? if you don't like factoring like what Jdoe did, you can use quadratic formula to solve for cos t. The result will be the same.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@loser66 well because my professor specifically said not too lol
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