Решите неравенство f'x>0 f(x)=2sin(п/3-x/4)-(√3)/4x
F`(x)=2cos(π/3-x/4)*(-1/4)-√3/4=-1/2*cos(x/4-π/3)-√3/4 cos(π/3-x/4)=cos(x/4-π/3) четная -1/2*cos(x/4-π/3)-√3/4>0 1/2*cos(x/4-π/3)<-√3/4<br>cos(x/4-π/3)<-√3/2<br>5π/6+2πk7π/6+2πk14π/3+8πk<x<6π+8πk,k∈z