lg khó

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niemkieuloveahbu

[TEX]Sin^{8}x - 2Sin^{4}x.Cos4x + 1 = 0\\ \Leftrightarrow sin^8x-2sin^4xcos4x+cos^24x+1-cos^24x=0 \\ \Leftrightarrow (sin^4-cos4x)^2+sin^24x=0\\ \{sin^4x=cos4x\\{sin4x=0} \Leftrightarrow x=\frac{\pi}{2}+k\pi[/TEX]
 
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