What is x if cos4x=cos2x ?
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We have to find x for which cos 4x = cos 2x
we use the relation cos 2x = 2(cos x)^2 - 1
cos 4x = cos 2x
=> 2(cos 2x)^2 - 1= cos 2x
=> 2(cos 2x)^2 - cos 2x - 1 = 0
=> 2(cos 2x)^2 - 2cos 2x + cos 2x - 1 = 0
=> 2cos 2x( cos 2x - 1) + 1( cos 2x - 1) = 0
=> ( 2cos 2x + 1)( cos 2x - 1) = 0
( 2cos 2x + 1) = 0
=> cos 2x = -1/2
=> 2x = 120 degrees
=> x = 60 degrees
cos 2x - 1 = 0
=> cos 2x = 1
=> x = 0
Therefore x = 0 + n*360 degrees and 60+ n*360 degrees
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To solve this equation, we'll move all the terms to the left side:
cos 4x - cos 2x = 0
Since the trigonometric functions are matching , we'll transform the difference into a product.
2 sin[(4x+2x)/2]*sin [(2x-4x)/2]=0
2 sin3x* sin (-x)=0
From this product of 2 factors, one or the other factor is zero.
sin3x=0
This is an elementary equation:
3x = (-1)^k*arcsin 0 + k*pi
3x = k*pi
x = k*pi/3, where k is an integer number.
We'll solve the second elementary equation:
sin (-x)=0
-sin (x)=0
sin x=0
x=(-1)^k arcsin 0 + k*pi
x=k*pi
The set of solutions is : S={k*pi/3}U{k*pi}.
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