## 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??

1. anonymous

@jdoe0001 I wrote it a little wrong!!

2. jdoe0001

well... your pi/3 is correct and your 5pi/3 is also correct so.... notice your range is from $$\bf cos(t)+1=2sin^2(t)\implies cos(t)+1=2[{\color{brown}{ 1-cos^2(t)}}] \\ \quad \\ cos(t)+1=2-2cos^2(t)\implies 2cos^2(t)+cos(t)-1=0 \\ \quad \\ \textit{factoring, we get}\\ \quad \\ [2cos(t)-1][cos(t)+1]=0\implies \begin{cases} 2cos(t)-1=0\\ 2cos(t)=1\\ cos(t)=\frac{1}{2}\\ t=cos^{-1}\left( \frac{1}{2} \right) \\\hline\\ cos(t)+1=0\\ cos(t)=-1\\ t=cos^{-1}(-1) \end{cases}$$

3. jdoe0001

cosine is -1 at $$\pi$$ thus $$\large \measuredangle t= \begin{cases} \frac{\pi }{3}\\ \pi \\ \frac{5\pi }{3} \end{cases}$$

4. anonymous

@jdoe0001 But we cant use quadratic formula tho?

5. 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.

6. anonymous

@loser66 well because my professor specifically said not too lol