Đáp án:Ta có tính chất $có^2\alpha+sin^2\alpha=1$
Giải thích các bước giải:
$a)2cos^4\alpha-sin^4\alpha+sin^2\alpha.cos^2\alpha+3sin^2\alpha$
$=2cos^4\alpha-sin^4\alpha+sin^2\alpha.cos^2\alpha+3sin^2\alpha(sin^2\alpha+cos^2\alpha)$
$=2cos^4\alpha-sin^4\alpha+sin^2\alpha.cos^2\alpha+3sin^4\alpha+3sin^2\alpha.cos^2\alpha$
$=2cos^4\alpha+4sin^2\alpha.cos^2\alpha+2sin^4\alpha$
$=2(sin^2\alpha+cos^2\alpha)^2=2$