[toán 11] Luong giac 11

N

niemkieuloveahbu

[TEX]\blue \text 3cosx + cos2x - cos3x + 1 = 2sinxcos3x\\ \Leftrightarrow cosx(3+2cosx)=(2sinx+1)cosx(4cos^2x-3)\\\Leftrightarrow \left[sinx=-1\\cosx=0\right.[/TEX]
 
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