sin3x+sin5x+2sin^2x/2=1

Question

sin3x+sin5x+2sin^2x/2=1

Answer 1

Как то так)))

sin3x+sin5x+2sin²(x/2)=1
2sin4x•cosx=1-2sin²(x/2)
2sin4x•cosx=cosx
cosx•(2sin4x-1)=0
1) cosx=0  =>   x=π/2+πn
2) sin4x=½  =>  4x=(-1)^n•π/6+πn  =>  x=(-1)^n•π/24+(π/4)•n