# Cos2x=cosx

What are x values for cos 2x = cos x?

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Here we have to find x for which cos 2x = cos x.

We know that cos 2x = 2(cos x)^2 - 1.

Therefore we can write the expression given as 2(cos x)^2 - 1 = cos x

or 2(cos x)^2 - 1 - cos x = 0

=> 2(cos x)^2 - 2 cos x + cos x - 1 = 0

=> 2cos x ( cos x -1 ) + 1 ( cos x -1 ) = 0

=> (2cos x +1)( cos x -1) = 0

=> 2cos x +1 = 0 or cos x -1 =0

=> cos x = -1/2 or cos x = 1

if cos x = -1/2 , we have x = 2*pi/3 + 2*n*pi or 2*n*pi - 2*pi/3

if cos x = 1 we have x = 0 or 2*n*pi

**Therefore the values of x for which cos 2x = cos x, are 2*pi/3 + 2*n*pi, 2*n*pi - 2*pi/3 , 0 and 2*n*pi, where n is any integer.**

Here is a similar problem:

https://www.youtube.com/watch?v=gx6OSGO80EI

To solve for x. cos2x = cosx

cos2x = cos(x+x) = cos^2x-sin^2x

cos2x = cos^2x-(1-cos^2x)

cos2x = 2cos^2x-1.

Substituting cos2x = 2cos^2x -1 in the give equation , we get:

2cos^2x-1 = cosx.

2cos^2x-cosx-1 = 0.

We factorise the left treating the equation as a quadratic in cosx.First we regroup the middle term by splitting:

2cos^2x-2cosx +cosx -1 = 0.

2cosx(cosx-1) +1(cosx-1) = 0

(2cosx +1)(cosx -1) = 0

Therefore 2cosx +1 = 0 Or cosx -1 = 0

2cosx+1 = 0 gives cosx = -1/2 . Or x = 2npi +or- 2pi/3.

cosx = 1 gives x = 0 or 2npi.

Therefore x= 2npi, or x = 2npi +2pi/3 or x = 2npi-2pi/3 n= 0,1,2....

We'll subtract cos x both sides and we'll get:

cos 2x - cos x = 0

Now, we'll apply the formula for the double angle 2x:

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

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

We'll write (sin x)^2 = 1 - (cos x)^2 (fundamental formula of trigonometry).

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

We'll remove the brackets:

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

We'll combine like terms:

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

We'll re-write the equation, substituting cos 2x y the expression (1).

2(cos x)^2 - 1 - cos x = 0

Now , we'll use substitution technique to solve the equation.

We'll note cos x = t and we'll re-write the equation in t:

2t^2 - t - 1 = 0

Since it is a quadratic, we'll apply the quadratic formula:

t1 = {-(-1) + sqrt[(-1)^2 + 4*2*1]}/2*2

t1 = [1+sqrt(1+8)]/4

t1 = (3+1)/4

t1 = 1

t2 = (1-3)/4

t2 = -1/2

Now, we'll put cos x = t1.

cos x = 1

Since it is an elementary equation, we'll apply the formula:

cos x = a

x = +/- arccos a + 2k*pi

x = arccos 1 + 2k*pi

x = 0

x = 2pi

cos x = t2

cos x = -1/2

x = pi - pi/3

x = 2pi/3

x = pi + pi/3

x = 4pi/3

**The solutions for the equation are:{0 ; 2pi/3 ; 4pi/3 ; 2pi}.**