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)'=6cos2x
f'(x)=0
6cos2x=0
2x=π/2+πk;x=π/4+πk/2
b)f'(x)=(4cos2x)'=4•(-sin2x)•(2x)'=-8sin2x
f'(x)=0
-8sin2x=0
2x=πk
x=πk/2;k€Z