(sin5x+sin3x) +sin2*4x =0 ;
2sin(5x+3x)/2*cos(5x-3x)/2 +2sin4x*cos4x=0 ;
2sin4x(cosx +cos4x) =0 ;
a) sin4x =0 ;
4x =π*k , k∈Z;
x = π/4 ,k∈Z .
b) cos4x +cosx =0 ;
2cos5x/2*cos3x/2 = 0;
[cos5x/2 =0 ;cos3x/2 =0.
[5x/2 =π/2 +π*k ; 3x/2 =π/2 +π*k .
[x = π/5 + 2π/5*k ; x = π/3 + 2π/3*k ; k∈Z .
ответ: π*k ; π/3 + 2π/3*k ; π/5 + 2π/5*k ; k∈Z .