Ask your own question, for FREE!
Mathematics 15 Online
OpenStudy (sleepyjess):

Find all solutions in the interval \(\sf [0, 2\pi).\) \(\sf cos^2x + 2 cos x + 1 = 0\)

OpenStudy (sleepyjess):

@amistre64

OpenStudy (anonymous):

\[\cos^2x+2\cos x+1=(\cos x+1)^2=0\]

OpenStudy (sleepyjess):

@sleepyhead314

OpenStudy (sleepyjess):

I just don't get it.......

OpenStudy (sleepyhead314):

I agree with siths imagine (random letter) X = (cos(x)) then X^2 +2X + 1 = 0 factor it, what do you get? :)

OpenStudy (sleepyhead314):

x^2 + 2x + 1 factor, what do you get? ^_^ this is Algebra 1 style :P

OpenStudy (sleepyjess):

(x+1)(x+1)

OpenStudy (sleepyhead314):

yes! which is the same as (x+1)^2 right? :)

OpenStudy (sleepyjess):

yes

OpenStudy (sleepyhead314):

so before I said let X = cos(x) now we will plug that back in :) so ( cos(x) + 1 )^2 <---- do you see where siths got that now? :)

OpenStudy (sleepyjess):

Yeah

OpenStudy (sleepyhead314):

then if we squareroot both sides we will still get cos(x) + 1 = 0 then subtract 1 from both sides

OpenStudy (sleepyjess):

cos(x)=-1

OpenStudy (sleepyhead314):

now when does cos(x) = -1 ?|dw:1418693093268:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!