Lời giải:Ta có \(f\left( x \right) = {\sin ^4}x + {\cos ^4}x = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x\)
\( = 1 - \frac{1}{2}{\sin ^2}2x = 1 - \frac{1}{4}\left( {1 - \cos 4x} \right) = \frac{3}{4} + \frac{1}{4}\cos 4x \Rightarrow f'\left( x \right) = - \sin 4x\)
Ta có \(\,g\left( x \right) = {\sin ^6}x + {\cos ^2}x = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^3} - 3{\sin ^2}x{\cos ^2}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\)
\( = 1 - \frac{3}{4}{\sin ^2}2x = 1 - \frac{3}{8}\left( {1 - \cos 4x} \right) = \frac{5}{8} + \frac{3}{8}\cos 4x \Rightarrow g'\left( x \right) = - \frac{3}{2}\sin 4x\)
Do đó \(3f'\left( x \right) - 2g'\left( x \right) + 2 = 3.\left( { - \sin 4x} \right) - 2\left( { - \frac{3}{2}\sin 4x} \right) + 2 = 2\).
