Gọi $3sin^4x+cos^4x=\frac{3}{4} (1)$$sin^4x+3cos^4x=A (12)$
Lấy $(12)-(1)=2cos^4x-2sin^4x=A-\frac{3}{4}$
$\Rightarrow cos2x=\frac{A}{2}-\frac{3}{8}$
Ta lại lấy $(12)+(1)=4(sin^2x+cos^2x)^2-8sin^2x.cos^2x=3+1-2sin^22x=2cos^22x+2=\frac{3}{4}+A$
Từ đó, suy ra: $2(\frac{A}{2}-\frac{3}{8})^2+2=\frac{3}{4}+A$
$\Leftrightarrow \frac{A^2}{2}-\frac{7A}{3}+\frac{49}{32}=0\Leftrightarrow A=\frac{7}{4}$