3 sin^4 - cos^4=1/2 tính 2 sin ^4 -cos^4

Đáp án: 1/4

Giải thích các bước giải:

$\begin{array}{l}
3{\sin ^4}x - {\cos ^4}x = \frac{1}{2}\\
 \Rightarrow 2{\sin ^4}x + {\sin ^4}x - {\cos ^4}x = \frac{1}{2}\\
 \Rightarrow \frac{1}{2}.4{\sin ^4}x + \left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^2}x - {{\cos }^2}x} \right) = \frac{1}{2}\\
 \Rightarrow \frac{1}{2}.{\left( {2{{\sin }^2}x} \right)^2} - 1.\left( {{{\cos }^2}x - {{\sin }^2}x} \right) = \frac{1}{2}\\
 \Rightarrow \frac{1}{2}.{\left( {1 - \cos 2x} \right)^2} - \cos 2x = \frac{1}{2}\\
 \Rightarrow \frac{1}{2}.{\cos ^2}2x - \cos 2x + \frac{1}{2} - \cos 2x = \frac{1}{2}\\
 \Rightarrow \frac{1}{2}{\cos ^2}2x - 2\cos 2x = 0\\
 \Rightarrow \cos 2x = 0\left( {do:\cos 2x \le 1\forall x} \right)\\
 \Rightarrow 2{\sin ^4}x - {\cos ^4}x\\
 = {\sin ^4}x + {\sin ^4}x - {\cos ^4}x\\
 = \frac{1}{4}.{\left( {2{{\sin }^2}x} \right)^2} - \cos 2x\\
 = \frac{1}{4}.{\left( {1 - \cos 2x} \right)^2} - 0\\
 = \frac{1}{4}
\end{array}$