Find all solutions of the equation in the range [0;pi]. 1+cosx+cos2x=0

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll start by re-writing the last term of the equation:

cos 2x = cos (x+x)

cos 2x = cos x*cos x - sin x*sin x

cos 2x = (cos x)^2 - (sin x)^2

We'll substitute 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 substitute 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 set each factor as zero:

cos x = 0

x = +/-arccos 0

x = +/- pi/2

Since the range of admissible values is [0 ; pi], we'll reject the solution -pi/2.

We'll put the other factor as zero:

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 range [0 ; pi], are {pi/2 ; 2pi/3}.

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