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Find all solutions in the interval \(\sf [0, 2\pi).\) \(\sf cos^2x + 2 cos x + 1 = 0\)
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@amistre64
\[\cos^2x+2\cos x+1=(\cos x+1)^2=0\]
@sleepyhead314
I just don't get it.......
I agree with siths imagine (random letter) X = (cos(x)) then X^2 +2X + 1 = 0 factor it, what do you get? :)
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x^2 + 2x + 1 factor, what do you get? ^_^ this is Algebra 1 style :P
(x+1)(x+1)
yes! which is the same as (x+1)^2 right? :)
yes
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? :)
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Yeah
then if we squareroot both sides we will still get cos(x) + 1 = 0 then subtract 1 from both sides
cos(x)=-1
now when does cos(x) = -1 ?|dw:1418693093268:dw|
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