SOLUTION: Having some trouble with this problem, any help is appreciated. Solve the following on the interval [0,2pi]: {{{2(cos(t))^2-cos(t)-1=0}}}

Algebra ->  Trigonometry-basics -> SOLUTION: Having some trouble with this problem, any help is appreciated. Solve the following on the interval [0,2pi]: {{{2(cos(t))^2-cos(t)-1=0}}}      Log On


   

Question 976076: Having some trouble with this problem, any help is appreciated.
Solve the following on the interval [0,2pi]:
2%28cos%28t%29%29%5E2-cos%28t%29-1=0
Answer by FrankM(1040) About Me  (Show Source):
You can put this solution on YOUR website!
2%28cos%28t%29%29%5E2-cos%28t%29-1=0
Think for a moment that this is similar to
2%28X%29%5E2-X-1=0
Which factors to
%282x%2B1%29%28X-1%29
And X=1 or x= -1/2
arcCos(1) is 0 and 2pi
arcCos -1/2 is 2/3 pi or 4/3 pi