Question

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??

Answer 1

@jdoe0001 I wrote it a little wrong!!

Answer 2

well... your pi/3 is correct and your 5pi/3 is also correct

so.... notice your range is from \(\bf cos(t)+1=2sin^2(t)\implies cos(t)+1=2[{\color{brown}{ 1-cos^2(t)}}]
\\ \quad \\
cos(t)+1=2-2cos^2(t)\implies 2cos^2(t)+cos(t)-1=0
\\ \quad \\
\textit{factoring, we get}\\ \quad \\ [2cos(t)-1][cos(t)+1]=0\implies
\begin{cases}
2cos(t)-1=0\\
2cos(t)=1\\
cos(t)=\frac{1}{2}\\
t=cos^{-1}\left( \frac{1}{2} \right)
\\\hline\\
cos(t)+1=0\\
cos(t)=-1\\
t=cos^{-1}(-1)
\end{cases}\)

Answer 3

cosine is -1 at \(\pi\)

thus \(\large \measuredangle t=
\begin{cases}
\frac{\pi }{3}\\
\pi \\
\frac{5\pi }{3}
\end{cases}\)

Answer 4

@jdoe0001 But we cant use quadratic formula tho?

Answer 5

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.

Answer 6

@loser66 well because my professor specifically said not too lol