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**