Nếu biết \(3{\sin ^4}x + 2{\cos ^4}x = \dfrac{{98}...

D

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

Ta có

\(\begin{array}{l}\,\,\,\,\,\,3{\sin ^4}x + 2{\cos ^4}x = \dfrac{{98}}{{81}}\\ \Leftrightarrow 2{\sin ^4}x + 3{\cos ^4}x + {\sin ^4}x - {\cos ^4}x = \dfrac{{98}}{{81}}\\ \Leftrightarrow A = \dfrac{{98}}{{81}} - \left( {{{\sin }^4}x - {{\cos }^4}x} \right)\end{array}\)

Ta có: \({\sin ^4}x - {\cos ^4}x = \left( {{{\sin }^2}x - {{\cos }^2}x} \right)\left( {{{\sin }^2}x + {{\cos }^2}x} \right) =  - \cos 2x\) 

\( \Rightarrow A = \dfrac{{98}}{{81}} + \cos 2x \Rightarrow \cos 2x = A - \dfrac{{98}}{{81}}\)   (1)

Ta lại có: \(3{\sin ^4}x + 2{\cos ^4}x + 2{\sin ^4}x + 3{\cos ^4}x = 5\left( {{{\sin }^4}x + {{\cos }^4}x} \right) = A + \dfrac{{98}}{{81}}\).

Mà \({\sin ^4}x + {\cos ^4}x = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x\)

\( = 1 - \dfrac{1}{2}{\sin ^2}2x = 1 - \dfrac{1}{2}\left( {1 - {{\cos }^2}2x} \right) = \dfrac{1}{2} + \dfrac{1}{2}{\cos ^2}2x\)

\( \Rightarrow 5\left( {\dfrac{1}{2} + \dfrac{1}{2}{{\cos }^2}2x} \right) = \dfrac{{98}}{{81}} + A\)   (2)

Thay (1) vào (2) ta có :

\(\begin{array}{l} \Rightarrow 5\left[ {\dfrac{1}{2} + \dfrac{1}{2}{{\left( {A - \dfrac{{98}}{{81}}} \right)}^2}} \right] = \dfrac{{98}}{{81}} + A\\ \Leftrightarrow \dfrac{5}{2}\left( {1 + {{\left( {A - \dfrac{{98}}{{81}}} \right)}^2}} \right) = A - \dfrac{{98}}{{81}} + \dfrac{{196}}{{81}}\end{array}\)

Đặt \(t = A - \dfrac{{98}}{{81}}\) ta có \(\dfrac{5}{2}\left( {1 + {t^2}} \right) = t + \dfrac{{196}}{{81}} \Leftrightarrow \left[ \begin{array}{l}t = \dfrac{{13}}{{45}}\\t = \dfrac{1}{9}\end{array} \right.\)

Với \(t = \dfrac{{13}}{{45}} \Rightarrow A = \dfrac{{13}}{{45}} + \dfrac{{98}}{{81}} = \dfrac{{607}}{{405}}\)

Với \(t = \dfrac{1}{9} \Rightarrow A = \dfrac{1}{9} + \dfrac{{98}}{{81}} = \dfrac{{107}}{{81}}\).

Chọn D