cos4x = cos2x

We know that:

cos2x = 1-2cos^2 x

==> cos4x = 1-2cos^2 (2x)

Now substitute:

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

==> 2cos^2 (2x) + cos2x - 1 = 0

Now factor:

==> (2cos2x -1)(cos2x + 1) = 0

==> 2cos2x -1 = 0 ==> cos2x =1/2 ==> 2x= pi/3

==> **x1= pi/6 , 7pi/6**

**==> x1= pi/6 + 2npi**

**==> x2= 7pi/6 + 2npi**

==> cos2x = -1 ==> 2x= pi ==> **x3 = pi/2 + 2npi.**

**==> x= { pi/6+2npi, 7pi/6+2npi, pi/2+2npi}**

For this question we use the relation:

cos A - cos B = 2 sin [(A + B)/2]*sin [(A-B)/2]=0

From cos 4x = cos 2x

=> cos 4x - cos 2x = 0

=> 2 sin [(4x + 2x) / 2 ]*sin [(4x - 2x ) / 2] = 0

=> sin [(4x + 2x) / 2 ]*sin [(4x - 2x ) / 2] = 0

=> sin (6x / 2 )*sin (2x / 2) = 0

=> sin 3x *sin x = 0

Now if sin 3x = 0 : 3x= k*pi

x= (1/3)*k*pi

If sin x = 0: x = k*pi

**Therefore x= k*pi or (1/3)*k*pi**

To detrmine x .

cos4x = cos2x.

We know that cos4a = 2cos^2x-1

So the given equation becomes:

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

2y^2 -y-1 = 0, where y = cos2x.

(2y+1 )(y-1) = 0.

Or y = 1 or 2y = -1/2.

Therefore y = 1 or cos2x = 1 which gives 2x = 2npi . Or x = npi.

y = -1/2 orcos2x = -1/2 gives 2x = 2npi+ or -2pi/3 . So x= (6n+or-2pi/3, n= 0, 1,2..

We'll re-write the identity, moving all terms to the left side. For this reason, we'll subtract cos 2x both sides:

cos4x - cos2x = 0

To solve the equation, we'll have to transform the difference of 2 like trigonometric functions into a product.

2 sin [(4x+2x)/2]*sin [(4x-2x)/2]=0

2 sin 3x * sin x=0

We'll divide by 2:

sin 3x * sin x=0

We'll set each factor from the product as 0.

sin 3x = 0

This is an elementary equation:

3x = (-1)^k*arcsin 0 + k*pi

3x = k*pi

We'll divide by 3 both sides:

**x = k*pi/3**

We'll solve the second elementary equation:

sin x=0

x = (-1)^k*arcsin 0 + k*pi

x = 0 + k*pi

**x = k*pi**

The set of solutions:

**S={k*pi/3}or{k*pi}**