Câu b) $PT \Leftrightarrow sin^6x+cos^6x=cos^2x-sin^2x+\frac{1}{16}$$\Leftrightarrow sin^2x(1+sin^4)+cos^2x(cos^4x-1)=\frac{1}{16}$
$\Leftrightarrow sin^2x(1+sin^4)+cos^2x(cos^2x-1)(cos^2+1)=\frac{1}{16}$
$\Leftrightarrow sin^2x(1+sin^4)-cos^2x.sin^2x.(cos^2x+1)=\frac{1}{16}$
$\Leftrightarrow sin^2x(1+sin^4x-cos^4x-cos^2x)=\frac{1}{16}$
$\Leftrightarrow sin^2x[(1-cos^2x)+(sin^2x+cos^2x)(sin^2x-cos^2x)]=\frac{1}{16}$
$\Leftrightarrow sin^2x(sin^2x+sin^2x-1+sin^2x)=\frac{1}{16}$
$\Leftrightarrow 3sin^4x-sin^2x-\frac{1}{16}=0$
Đến đây bạn tự giải nha