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

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}.