#cos2(x-pi/4) = cos(2x-pi/2) = sin2x#

(If you don't remember the identity above, use the difference formula for

#cos(2x-pi/2) = cos(2x)cos(pi/2)+sin(2x)sin(pi/2) = sin(2x)#)

So #f(x) = 3sin(2x)+1#

and #f'(x) = 6cos(2x)# which exists for all #x# and is #0# when

#cos(2x) = 0#. Which is true exactly when

#2x = pi/2 + pik# for integer #k#. Or,

#x = pi/4 + pi/2k# for integer #k#.

**Without rewriting** we can get the same answer (of course).

#f(x) = 3cos2(x-pi/4)+1#

#f'(x) = -3sin2(x-pi/4)*[d/dx(2(x-pi/4)] = -3sin2(x-pi/4)*[2(1)]#

#f'(x) = -6sin2(x-pi/4)#

F'(x) is never undefined and it is #0# for

#sin2(x-pi/4) = 0# Which is true exactly when

#2(x-pi/4) = pik# for integer #k#. Or,

#x -pi/4= pi/2k# for integer #k#. So we need

#x = pi/4+ pi/2k# for integer #k#.