Решите уравнение f ' (x) = 0
a) f (x) = 3 sin 2 x
б) f (x) = 4 cos 2 x

а)f'(x)=(3sin2x)'=3•cos2x*(2x)'=6cos2xf'(x)=06cos2x=02x=π/2+πk;x=π/4+πk/2b)f'(x)=(4cos2x)'=4•(-sin2x)•(2x)'=-8sin2xf'(x)=0-8sin2x=02x=πkx=πk/2;k€Z