SOLUTION: solve for the exact value of x where 0 greater than or equal to X and less than/equal to 2 pi. cos^2 x-1=0 I did the factoring and FOILing and The two answers I got were:

Algebra ->  Trigonometry-basics -> SOLUTION: solve for the exact value of x where 0 greater than or equal to X and less than/equal to 2 pi. cos^2 x-1=0 I did the factoring and FOILing and The two answers I got were:       Log On


   

Question 206572: solve for the exact value of x where 0 greater than or equal to X and less than/equal to 2 pi.
cos^2 x-1=0
I did the factoring and FOILing and
The two answers I got were:
cosx=-1 cosx=1
x=pi x=0,2pi
According to my instructor cosx=1's answers are not completly correct. I dont see what the problem is since cos(0) and cos(2pi) both =1. Am I missing something here? Thank you for your help.
Answer by mickclns(59) About Me  (Show Source):
You can put this solution on YOUR website!
I'd be interested in seeing your instructor's answer (please email me). It seems correct to me, too.
Things I can see:
You say "where 0 greater than or equal to X and less than/equal to 2 pi" by which I assume you mean 0%3C=x%3C=2pi although, strictly speaking, what you have written is 0%3E=x%3C=2pi which is meaningless or if it could be given meaning there would be no solution possible because there is no x such that 0 greater than or equal to x and (x is) less than/equal to 2 pi.
Another possibility, the span usually given in problems such as this is 0%3C=x%3C2pi. So I would check to see whether the second really includes the =. If so, maybe the instructor intended it not to have the = so 2pi would not be an answer. So take another look and ask again.