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

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### 2 Answers

We have to solve cos 3x = cos x for x in the interval (0 , 2*pi)

We know that cos 3x = 4(cos x)^3 - 3cos x

cos 3x = cos x

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

let cos x = t

=> 4t^3 - 4t = 0

=> (4t( t^2 - 1) = 0

=> 4t*(t -1)(t + 1) = 0

=> t = 0 and t = 1 and t = -1

cos x = 0 => x = pi/2 and 3*pi/2

cos x = 1 => x = 0, 2*pi but this does not lie in the given interval

cos x = -1 => x = pi

**Therefore x can take the values pi/2 , pi and 3*pi/2**

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}.**