Как решать:
5sin2x=0
-4cos(p/4-0,1x)=0

5sin(2x)=0, sin(2x)=0,  2x = arc sin 0 = πk, k ∈ Z.x = (πk/2), k ∈ Z.-4cos((π/4)-0,1x)=0,  cos((π/4)-0,1x)=0,  (π/4)-0,1x = arc cos 0. (π/4)-0,1x = (π/2) + πk,  k ∈ Z.x = ((π/4)-(π/2)-πk)*10 = (-10π/4) - 10πk = (-5π/2) - 10π, k ∈ Z.