Determine x for cos4x = cos2x . Determine x for cos4x = cos2x .
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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}
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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}
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