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