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Determine the general formula(s) for the solution. Then, determine the specific solutions on the interval [0,2pi). 2cos^2(theta)-3cos(theta)+1=0

Trigonometry
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still need help?
yes i do :(
let cos\(\theta\)= x, you have the form of \(2x^2-3x+1=0\) solve for x as a quadratic equation, then plulg back to x = cos \(\theta\) to solve for \(\theta\)

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Other answers:

thank you! but i mistyped the question it is actually plus 3, not 1. :( so would the quadratic equation be \[2x ^{2}-3x+3\]?
i cannot figure out how to factor that
which class are you in? do you know Euler's formula for complex number?
i am just in a college trig class. 1316. we have not gotten to euler's formula thus far
this is the actual problem \[2\sin ^{2}\Theta-3\cos \Theta=-3\]
hey, it totally different,!! heehe , let see. 2 sin^2 = 2 - 2cos^2 therefore, 2-2cos^2 -3 cos +3 =0 2cos^2 +3 cos -5 =0, can solve it now
you should post the original one, friend.
you really scared me!! I thought I met a Ph.D student.
lol i know i got it confused with another problem. I was really good at algebra but this stuff is just killing me...
hahaha no ph.d student here
so? everything is quite simple, right?
so i have \[2x ^{2}+3x-5=0\] and now factor? im sorry this class has confused me so much
if you know how to factor, do it. if not, use discriminant. Either way is ok.
x=-5/2 and x=1?
yes, and reject -5/2, out of interval
okay. so: cos(theta)=1? how do i make the general formula in form of of: _+_k
not finish yet, cos \(\theta\) =1 \(\rightarrow \theta\) = 0 + 2k\(\pi\) that 's the general formula solution for you
so the only solution is 0? because subbing 1 for k would make 2pi which would not be included in the interval?
yes, you are right. general form is 2kpi, specific solution in the interval is 0 Medal for a good student
oh my gosh thank you so much for your help!
do I help? No! just you work on it, I did nothing. hehee
:) :)

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