cosx+cos5x+2sin²x=12cos3x•cos2x=cos²x-sin²x2cos3x•cos2x-cos2x=0cos2x•(2cos3x-1)=01) cos2x=0 => 2x=π/2+πn => x=π/4+(π/2)•n 2) cos3x=½ => 3x=±π/3+2πn => x=±π/9+(2π/3)•n
Вроде бы так)
cosx+cos5x+2sin²x=1
2cos3x•cos2x=cos²x-sin²x
2cos3x•cos2x-cos2x=0
cos2x•(2cos3x-1)=0
1) cos2x=0 => 2x=π/2+πn => x=π/4+(π/2)•n
2) cos3x=½ => 3x=±π/3+2πn => x=±π/9+(2π/3)•n
Вроде бы так)