## 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. jdoe0001

how did you get pi/3 and 5pi/3?

2. anonymous

I basically equaled it to zero and solved from there

3. jdoe0001

equalled to 0? $$\bf cos(t)=2sin^2(t)\implies cos(t)-2sin^2(t)=0?$$

4. anonymous

Yepp thats what I did except its 2sin squared t - cos t=0

5. jdoe0001

hmm I don't see how pi/3 and 5pi/3 comes out of that though

6. anonymous

@jdoe0001 cos theta of 1/2 equals to pi/3 or cos =-1 which is 5pi/3

7. jdoe0001

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

8. anonymous

We can't use quadtric formula tho :///

9. anonymous

@jdoe0001

10. jdoe0001

can't? if it's a quadratic one.. why not?

11. anonymous

@jdoe0001 cuz my professor only said to use trig functions which is where I am running into problems

12. jdoe0001

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

13. anonymous

@jdoe0001 Yeah it is I forgot to put the one!

14. jdoe0001

well.. you may want to repost with the proper expression then, so we can see it