anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@jdoe0001 I wrote it a little wrong!!
jdoe0001
  • jdoe0001
well... your pi/3 is correct and your 5pi/3 is also correct so.... notice your range is from \(\bf cos(t)+1=2sin^2(t)\implies cos(t)+1=2[{\color{brown}{ 1-cos^2(t)}}] \\ \quad \\ cos(t)+1=2-2cos^2(t)\implies 2cos^2(t)+cos(t)-1=0 \\ \quad \\ \textit{factoring, we get}\\ \quad \\ [2cos(t)-1][cos(t)+1]=0\implies \begin{cases} 2cos(t)-1=0\\ 2cos(t)=1\\ cos(t)=\frac{1}{2}\\ t=cos^{-1}\left( \frac{1}{2} \right) \\\hline\\ cos(t)+1=0\\ cos(t)=-1\\ t=cos^{-1}(-1) \end{cases}\)
jdoe0001
  • jdoe0001
cosine is -1 at \(\pi\) thus \(\large \measuredangle t= \begin{cases} \frac{\pi }{3}\\ \pi \\ \frac{5\pi }{3} \end{cases}\)

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anonymous
  • anonymous
@jdoe0001 But we cant use quadratic formula tho?
Loser66
  • Loser66
why 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
  • anonymous
@loser66 well because my professor specifically said not too lol

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