Question

решите неравенство f'x>0

f(x)=2sin(п/3-x/4)-(√3)/4x

Answer 1

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
cos(x/4-π/3)<-√3/2
5π/6+2πk<x/4-π/3<7π/6+2πk
7π/6+2πk<x/4<3π/2+2πk
14π/3+8πk<x<6π+8πk,k∈z