Найдите производную фунуции f(x)=1/3 sin3x - 1/2 x и решите неравенство f"(x)< 0

f`(x)=(1/3 sin3x - 1/2 x)`=(1/3)·(sin3x)`-(1/2)·(x)`=
=(1/3)·(cos3x)·(3x)`-(1/2)=
=cos3x-(1/2)
f``(x)=(cos3x -(1/2))`=(cos3x)`-(1/2)`=-sin3x·(3x)`-0=-3sin3x

f"(x)<0
-3sin3x <0
sin3x >0
0+2πk < 3x < π+2πk, k∈Z
(2π/3)·k < x < (π/3) + (2π/3)·k, k∈Z

О т в е т. ((2π/3)·k; (π/3) + (2π/3)·k), k∈Z.