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helpplease1
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
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 \(\theta\)
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.
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