To find the critical points of the function ( f(x) = \cos(2x) - 2\cos(x) ), we need to find where the derivative is equal to zero or does not exist.

First, let's find the derivative of the function ( f(x) ):

( f'(x) = -2\sin(2x) + 2\sin(x) )

Next, set ( f'(x) = 0 ) and solve for x:

( -2\sin(2x) + 2\sin(x) = 0 )

( 2\sin(x)(\cos(x) - 1) = 0 )

This equation is satisfied when ( \sin(x) = 0 ) or ( \cos(x) = 1 ).

When ( \sin(x) = 0 ), x can be an integer multiple of ( \pi ): x = n( \pi ), where n is an integer.

When ( \cos(x) = 1 ), x = 2( \pi ).

Therefore, the critical points of the function occur at x = n( \pi ), where n is an integer, and x = 2( \pi ).

If you need more help, feel free to ask!