Как то так)))
sin3x+sin5x+2sin²(x/2)=12sin4x•cosx=1-2sin²(x/2)2sin4x•cosx=cosxcosx•(2sin4x-1)=01) cosx=0 => x=π/2+πn2) sin4x=½ => 4x=(-1)^n•π/6+πn => x=(-1)^n•π/24+(π/4)•n
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Verified answer
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