K kittykat7 May 2010 7 0 May 25, 2010 #1 Please help me solve this problem.... it's been awhile so I am a bit rusty! Thanks cos x + 2 cos x = -1 over the interval [0, 2 p)

Please help me solve this problem.... it's been awhile so I am a bit rusty! Thanks cos x + 2 cos x = -1 over the interval [0, 2 p)

masters MHF Helper Jan 2008 2,550 1,187 Big Stone Gap, Virginia May 25, 2010 #2 kittykat7 said: Please help me solve this problem.... it's been awhile so I am a bit rusty! Thanks cos x + 2 cos x = -1 over the interval [0, 2 p) Click to expand... Hi kittykat7, I'm going to assume that you meant to write: \(\displaystyle \cos^2x+2 \cos x=-1\) In which case, \(\displaystyle \cos^2 x+2\cos x+1=0\) \(\displaystyle (\cos x+1)^2=0\) \(\displaystyle \cos x=-1\) \(\displaystyle x=\{\pi\}\) Reactions: kittykat7

kittykat7 said: Please help me solve this problem.... it's been awhile so I am a bit rusty! Thanks cos x + 2 cos x = -1 over the interval [0, 2 p) Click to expand... Hi kittykat7, I'm going to assume that you meant to write: \(\displaystyle \cos^2x+2 \cos x=-1\) In which case, \(\displaystyle \cos^2 x+2\cos x+1=0\) \(\displaystyle (\cos x+1)^2=0\) \(\displaystyle \cos x=-1\) \(\displaystyle x=\{\pi\}\)

K kittykat7 May 2010 7 0 May 25, 2010 #3 masters said: Hi kittykat7, I'm going to assume that you meant to write: \(\displaystyle \cos^2x+2 \cos x=-1\) In which case, \(\displaystyle \cos^2 x+2\cos x+1=0\) \(\displaystyle (\cos x+1)^2=0\) \(\displaystyle \cos x=-1\) \(\displaystyle x=\{\pi\}\) Click to expand... Thanks, I'm actually not sure what it was asking, that is how I received the question. That is one of the choices for an answer, so I am assuming that is what I was asking. Thank you for your speedy response!

masters said: Hi kittykat7, I'm going to assume that you meant to write: \(\displaystyle \cos^2x+2 \cos x=-1\) In which case, \(\displaystyle \cos^2 x+2\cos x+1=0\) \(\displaystyle (\cos x+1)^2=0\) \(\displaystyle \cos x=-1\) \(\displaystyle x=\{\pi\}\) Click to expand... Thanks, I'm actually not sure what it was asking, that is how I received the question. That is one of the choices for an answer, so I am assuming that is what I was asking. Thank you for your speedy response!