Giải thích các bước giải:
$A=2\cos^4x-\sin^4x+\sin^2x.\cos^2x+3\sin^2x\\
=2\cos^4x-\sin^4x+\sin^2x.(1-\sin^2x)+3\sin^2x\\
=2\cos^4x-\sin^4x+\sin^2x-\sin^4x+3\sin^2x\\
=2\cos^4x-2\sin^4x+4\sin^2x\\
=2(\cos^4x-\sin^4x)+4\sin^2x\\
=2(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)+4\sin^2x\\
=2\cos^2x-2\sin^2x+4\sin^2x\\
=2\cos^2x+2\sin^2x\\
=2(\cos^2x+\sin^2x)\\
=2$
