Đáp án:
\(\left[ \begin{array}{l}x=±\dfrac{π}{6}\\x=\dfrac{π}{2} + kπ\end{array} \right.\) , `k` ∈ `Z`
Giải thích các bước giải:
`cos^{4}``x` `-` `sin^{4}``x` + `cos4x` `=` `0`
⇔ ( `cos^{2}``x` `-` `sin^{2}``x` )( `cos^{2}``x` `+` `sin^{2}``x` ) `+` ( `2` `cos^{2}``2x` `-` `1` ) `=` `0`
⇔ `cos2x` `+` `2``cos^{2}` `2x` `-` `1` = `0`
⇔ \(\left[ \begin{array}{l}cos2x=\dfrac{π}{3}\\cos2x=-1\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}2x=±\dfrac{π}{3} + k2π \\2x=π + k2π\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=±\dfrac{π}{6}+kπ\\x=\dfrac{π}{2}+kπ\end{array} \right.\) , `k` ∈ `Z`
