Cho 3sin⁴x – cos⁴x = \(\frac{1}{2}.\) Tính A = 2sin⁴x – cos⁴x.

Ta có: sin²x + cos²x = 1

⇔ cos²x = 1 – sin²x

Mà 3sin⁴x – cos⁴x = \(\frac{1}{2}.\)

⇔ 3sin⁴x – (1 – sin²x)² = \(\frac{1}{2}.\)

⇔ 3sin⁴x – 1 + 2sin²x – sin⁴x – \(\frac{1}{2}\)= 0

⇔ 2sin⁴x + 2sin²x – \(\frac{3}{2}\) = 0

⇔ \(\left[ \begin{array}{l}{\sin ^2}x = \frac{1}{2}\\{\sin ^2}x = \frac{{ - 3}}{2}\left( L \right)\end{array} \right.\)

⇔ sin²x = \(\frac{1}{2}\)

⇔ cos²x = \(\frac{1}{2}\)

A = 2sin⁴x – cos⁴x = \(2.{\left( {\frac{1}{2}} \right)^2} - {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4}\).