SOLUTION: Solve for theta: cos(2theta+pi) = +/- (sqrt(2)/2) [There are no restrictions, so answer should have an infinite number of solutions.] A little bit of background... I realize

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Question 977063: Solve for theta: cos(2theta+pi) = +/- (sqrt(2)/2) [There are no restrictions, so answer should have an infinite number of solutions.]
A little bit of background... I realize this may be a cut-above regular algebra and as such I'm sorry for posting it on here. The thing is, it's actually really easy stuff and I already know how to solve these. What's confusing me right now is how to write the answer(s) in a way that shows there are an infinite number of solutions. Most teachers restrict the domain. Not mine! Thank you guys again, very much!
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
cos%282%2Atheta%2Bpi%29=0+%2B-+sqrt%282%29%2F2
The solutions are,
2%2Atheta%2Bpi=%281%2F4%29pi%2B2pi%2Ak
2%2Atheta%2Bpi=%283%2F4%29pi%2B2pi%2Ak
2%2Atheta%2Bpi=%285%2F4%29pi%2B2pi%2Ak
2%2Atheta%2Bpi=%287%2F4%29pi%2B2pi%2Ak
So then,
2%2Atheta=%281%2F4%29pi%2B2pi%2Ak-pi
2%2Atheta=%283%2F4%29pi%2B2pi%2Ak-pi
2%2Atheta=%285%2F4%29pi%2B2pi%2Ak-pi
2%2Atheta=%287%2F4%29pi%2B2pi%2Ak-pi
And,
theta=%281%2F8%29pi%2Bpi%2Ak-pi%2F2
theta=%283%2F8%29pi%2Bpi%2Ak-pi%2F2
theta=%285%2F8%29pi%2Bpi%2Ak-pi%2F2
theta=%287%2F8%29pi%2Bpi%2Ak-pi%2F2
Simplifying,
theta=pi%2A%28k-3%2F8%29
theta=pi%2A%28k-1%2F8%29
theta=pi%2A%28k%2B1%2F8%29
theta=pi%2A%28k%2B3%2F8%29
where k is any integer.