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Find all solutions in the interval [0, 2π).

cos2x + 2 cos x + 1 = 0

x = 2π

x = π

x = pi divided by four., seven pi divided by four.

x = pi divided by two., three pi divided by two.

Question

Medal, fan, testimonial for quickest answer. Please Help. 
Find all solutions in the interval [0, 2π).

cos2x + 2 cos x + 1 = 0 

 x = 2π 

 x = π 

 x = pi divided by four., seven pi divided by four.

 x = pi divided by two., three pi divided by two.

anonymous · Answer

$$\cos^2x + 2 \cos x+1$$

anonymous · Answer

@haleyelizabeth2017 @gottennis121

anonymous · Answer

@Dani_Rose

anonymous · Answer

Please help me @Dani_Rose @leticiau @Preetha

dani_rose · Answer

idk how to do this but maybe this will help https://www.youtube.com/watch?v=CnhmIbagc04

anonymous · Answer

@Jaynator495 @KendrickLamar2014 @Preetha

anonymous · Answer

@jabez177 @imqwerty

anonymous · Answer

@Marie1738 @Qwertty123

anonymous · Answer

Please Help Me! These are my options:
 x = 2π 

 x = π 

 x = pi divided by four., seven pi divided by four.

 x = pi divided by two., three pi divided by two.

anonymous · Answer

Im not looking for answer, I just need help solving

marie1738 · Answer

Okay. So I suck at math, but English. tag me anytime... Just some knowledge you should know beforehand..... BUT. I am very capable of doing research online, sooo.. Here are some explanations I found online in reference to this particural problem:


#1: Basically the solutions of this particular equation is going to be restricted to one revolution. 

So first: 

(cosx+1)^2 = cos^2(x)+2cosx+1=0 
cosx+1=0 
cosx = -1 
(cosines equal -1 when x=pi, and pi only occurs once from 0 to pi) 
x=pi 


#2: we need the trig identities to rescue us! 
cos(2x) = cos^2(x) - sin^2(x) = 2 cos^2(x) - 1 

here the 2nd does the trick: 
subbing in, we get 
2 cos^2(x) - 1 + 2cosx + 1=0 
or 2 cos^2(x) + 2cosx=0 
cos^2(x) + cosx = 0 
cosx(cosx+1)=0 

so cosx = 0 or cosx = -1 
this happens at pi/2 and 3pi/2

#3: you can factor as follows: 
(cos x + 1)^2 = 0 
cos x = -1 

This occurs when x = pi.

marie1738 · Answer

I do hope I helped. Even if I did not personally know the answer to the problem

anonymous · Answer

Thank you so much! Im not understanding much but I will look into it. I think it is saying that it happens at pi/2 and 3pi/2

marie1738 · Answer

I can't tell you, but the answer is in fact  B. x= pi

marie1738 · Answer

Have a great day young friend

anonymous · Answer

Thanks! You too!