Find all solutions of the equation in the range [0;pi] . Identify solutions of the equation in [0;pi] . 1 + cosx + cos2x = 0
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We have to solve 1 + cos x + cos 2x = 0 for x in the interval (0, pi)
1 + cos x + cos 2x = 0
=> 1 + cos x + 2(cos x)^2 - 1 = 0
=> cos x + 2(cos x)^2 = 0
=> cos x( cos x + 2) = 0
=> cos x = 0
cos x = -2 is not possible
The solution of the equation in the interval (0, pi) is x = pi/2 and x = 2*pi/3
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We'll apply the double angle identity:
cos 2x = (cos x)^2 - (sin x)^2
We'll replace the term (sin x)^2 by the difference 1 - (cos x)^2
cos 2x = (cos x)^2 - 1 + (cos x)^2
We'll combine like terms:
cos 2x = 2(cos x)^2 - 1
Now, we'll replace the last term cos 2x by it's expression:
1 + cos x + 2(cos x)^2 - 1 = 0
We'll eliminate like terms:
2(cos x)^2 + cos x = 0
We'll factorize by cos x:
cos x(2cos x + 1) = 0
We'll cancel each factor:
cos x = 0
x = +/-arccos 0
x = +/- pi/2
Since the interval of admissible values is [0 ; pi], we'll keep only the solution pi/2.
We'll cancel the next factor:
2cos x + 1 = 0
2cos x = -1
cos x = -1/2
x = pi - arccos (1/2)
x = pi - pi/3
x= 2pi/3
The solutions of the given equation, in the interval [0 ; pi], are {pi/2 ; 2pi/3}.
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