[tex]\sin^6x - 3 \sin^4x + 3 \sin^2x + \cos^6x + 1[/tex]
[tex]\sin^6x+\cos^6x=\left ( \sin^2x+\cos^2x \right )\left ( \sin^4x-\sin^2x\cos^2x+\cos^4x \right )=\\=\sin^4x-\sin^2x\cos^2x+\cos^4x=\sin^4x-\sin^2\left ( 1-\sin^2x \right )+\left ( \cos^2x \right )^2=\\=\sin^4x-\sin^2x+\sin^4x+\left ( 1-\sin^2x \right )^2=\\=2\sin^4x-\sin^2x+1-2\sin^2x+\sin^4x=\\3\sin^4x-3\sin^2x+1\\[/tex]
Подставляем в наше выражение
[tex]3\sin^4x-3\sin^2x+1-3\sin^4x+3\sin^2x+1=2[/tex]
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[tex]\sin^6x - 3 \sin^4x + 3 \sin^2x + \cos^6x + 1[/tex]
[tex]\sin^6x+\cos^6x=\left ( \sin^2x+\cos^2x \right )\left ( \sin^4x-\sin^2x\cos^2x+\cos^4x \right )=\\=\sin^4x-\sin^2x\cos^2x+\cos^4x=\sin^4x-\sin^2\left ( 1-\sin^2x \right )+\left ( \cos^2x \right )^2=\\=\sin^4x-\sin^2x+\sin^4x+\left ( 1-\sin^2x \right )^2=\\=2\sin^4x-\sin^2x+1-2\sin^2x+\sin^4x=\\3\sin^4x-3\sin^2x+1\\[/tex]
Подставляем в наше выражение
[tex]3\sin^4x-3\sin^2x+1-3\sin^4x+3\sin^2x+1=2[/tex]