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

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

Answer 2

I basically equaled it to zero and solved from there

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

@jdoe0001

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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