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Решите задачу:

$sin2x=\frac{2}{5}\\\\sin^2x+cos^2x=1\\\\(sin^2x+cos^2x)^2=1\; \; \to \; \; sin^4x+2sin^2xcos^2x+cos^4x=1\; \; \to \\\\sin^4x+cos^4x=1-2(sinxcosx)^2=1-2\cdot (\frac{1}{2}sin2x)^2=1-\frac{1}{2}\cdot (\frac{2}{5})^2=\\\\=1-\frac{2}{25}=\frac{23}{25}\\\\(sin^4x+cos^4x)^2=(\frac{23}{25})^2\; \; \to \\\\sin^8x+2sin^4xcos^4x+cos^8x=\frac{529}{625}\\\\sin^8x+cos^8x=\frac{529}{625}-2\cdot (sinxcosx)^4=$

$=\frac{529}{625}-\frac{1}{2}\cdot sin^42x=\frac{529}{625}-\frac{1}{2}\cdot (\frac{2}{5})^4=\frac{529}{625}-\frac{8}{625}=\frac{521}{625}$