#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#.
