# 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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### 2 Answers

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**

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}.