[toan11] Bài lượng giác đây.

K

keosuabeo_93

sinx +sin2x + sin3x = 1 + cosx + cos2x
=> 2sin2x.cosx+ sin 2x=cosx+2cos^2x
=> sin2x.(2cosx+1)- cosx.(1+2cosx)=0
=> (sin 2x- cos x)(2cosx+1)=0
 
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N

niemtin_267193

sinx +sin2x + sin3x = 1 + cosx + cos2x
------------------------Thanks mọi người trước nha!----------------------------------------------
[TEX]sinx + sin3x + sin2x = 1 + cos2x +cosx[/TEX]
[tex]\Leftrightarrow 2sin2x.cosx + sin2x = 1 + cosx + 2cos^2x - 1[/tex]
[tex]\Leftrightarrow sin2x.(2cosx +1) - cosx.(1 + 2cosx)=0[/tex]
[tex]\Leftrightarrow ( 2cosx +1)(sin2x - cosx)=0[/tex]
[TEX]\Leftrightarrow cosx=\frac{-1}{2}[/TEX] hoặc [TEX]sin2x=cosx[/TEX]
 
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