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Find all solutions to the equation. cos2x + 2 cos x + 1 = 0

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For cos(2x) * (2cos(x) + 1) = 0, use the double angle identity for cos(2x), which is cos^2 x - sin^2 x = cos^2 x - (1-cos^2) = 2cos^2 x - 1. So we have (2cos^2 x - 1)(2cos x + 1) = 0. So 2cos^2 x -1 = 0 or x = 0 and 2pi. For 2sec^2 x + tan^2 x - 3 = 0, use the identity sec^2 x = tan^2 x + 1, so we have 2(tan^2 x + 1) + tan^2 x - 3 = 0 or 2tan^2 x + tan^2 x - 1 = 0 or 3 tan^2 x = 1. So x = pi/2, pi/2 + pi = 3pi/2.
hi the first thing I tried to solve this problem was to factor it so I have (cos x+1)(cos x+1) but now I dont know what to next
Unclear how to proceed. What is the first term, \(\cos(2x)\) or \(\cos^{2}(x)\)?

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Other answers:

cos \[\cos^{2}x\]
Perfect. The we have only to solve - by factoring. \(\cos^{2}x + 2\cos x + 1 = 0\) \((\cos(x) + 1)^{2} = 0\) Now what?
That is where I got stuck I am not sure what to do after factoring
What's the point of factoring? Why do we do that at all? If a*b = 0, what do we know about a or b? To contract, if a*b = 4, what do we NOT know about either a or b?
what each of the equal?
Kind of. a*b = 0 tells us that either a = 0 or b = 0. What can we tell about a and b when it isn't zero? a*b = 4, for example? Is one of them zero? Aboslutely not, but that's about all we can tell. Is a = 4? Maybe. Is b = 12? Maybe. Not a clue. Only "=0" is particularly helpful. So, what should we do with that factored expression?
tkhunny your explanations are more complex than the problem itself :)
That's why we invent notation, so that problems can be simpler than explanations. If we have this: Factor1 * Factor2 = 0 Then we must have Either this Factor1 = 0 or this Factor2 = 0 How can we apply this to your factored equation?
cos x+1(0)=0 I am not sure
\((\cos(x)+1)^{2} = 0\) We have two factors, but hey are exactly the same. This limits the possible results. \(\cos(x) + 1 = 0\) \(\cos(x) = -1\) \(x = \dfrac{3\pi}{2} + 2k\pi\) where \(k\in \mathbb{Z}\) Do you know what all that means?
I do... but if cos(x)=-1 wouldn't [x=\pi\]
anyway I got it ... thanks for the help!
Awesome. How did I do that?! Right you are. \(x = \pi + 2k\pi\)

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