b) Ta có: \({\sin ^2}x + {\cos ^2}x = 1 \Leftrightarrow {\cos ^2}x = 1 - {\sin ^2}x\)
Thay vào ta được:
\(\begin{array}{l}3{\sin ^4}x - {\cos ^4}x = \dfrac{1}{2}\\ \Leftrightarrow 3{\sin ^4}x - {\left( {1 - {{\sin }^2}x} \right)^2} = \dfrac{1}{2}\\ \Leftrightarrow 3{\sin ^4}x - {\sin ^4}x + 2{\sin ^2}x - 1 = \dfrac{1}{2}\\ \Leftrightarrow 2{\sin ^4}x + 2{\sin ^2}x - \dfrac{3}{2} = 0\\ \Leftrightarrow \left[ \begin{array}{l}{\sin ^2}x = \dfrac{1}{2}\\{\sin ^2}x = - \dfrac{3}{2}\left( {Loai} \right)\end{array} \right.\\ \Rightarrow {\cos ^2}x = 1 - \dfrac{1}{2} = \dfrac{1}{2}\\ \Rightarrow B = {\sin ^4}x + 3{\cos ^4}x = {\left( {{{\sin }^2}x} \right)^2} + 3{\left( {{{\cos }^2}x} \right)^2}\\ = {\left( {\dfrac{1}{2}} \right)^2} + 3.{\left( {\dfrac{1}{2}} \right)^2} = 1\end{array}\)
