Solve the equation if x is in interval (0;2pi) cos3x=cosx

First of all, we'll try to calculate cos 3x = cos
(2x+x)

cos (2x+x) = cos 2x*cos x - sin x*sin
2x

sin (2x) = 2sin x*cos x

cos
2x = 2 (cos x)^2 - 1

cos 3x = [2 (cos x)^2 - 1]*cos x - 2[1
- (cos x)^2]*cos x

cos 3x = 4(cos x)^3 - 3cos
x

So the equation will
become:

4(cos x)^3 - 3cos x= cos
x

We'll move all terms to one
side:

4(cos x)^3 - 3cos x- cos x =
0

4(cos x)^3 - 4cos x =
0

We'll factorize by
4cosx:

4cos x*[(cos x)^2 - 1] =
0

4cos x = 0

cos x =
0

The values of the angle x, located in the range (0 ;
2pi), for the function cosine is 0, are: x = pi/2 and x =
-pi/2.

(cos x)^2 - 1 = 0

Since
it is a difference of squares, we'll re-write it as a
product:

(cos x - 1)(cos x + 1) 
=0

cos x - 1 = 0

cos x =
1

The values of the angle x, located in the range (0 ;
2pi), for the function cosine is 1, are: x = 0 and x
=2pi.

Neither of these values are located in the given
range, since the interval is opened both sides, meaning that the values 0 aand 2pi are
not included.

cos x + 1 =
0

cos x = -1

The value of the
angle x, located in the range (0 ; 2pi), for the function cosine is -1, is: x =
pi.

The solutions of the equation, located in
the interval (0 ; 2pi),are: {pi/2 ; pi ; 3pi/2}.