Giải phương trình sau: \({\sin ^8}x + co{s^8}x = \frac{1}{8}\).

Ta có:

\(\begin{array}{l}{\sin ^2}x + co{s^2}x = 1\\ \Rightarrow {\sin ^4}x + co{s^4}x = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x.co{s^2}x\\ = 1 - 2{\sin ^2}x.co{s^2}x\\ \Rightarrow {\sin ^8}x + co{s^8}x = {\left( {{{\sin }^4}x + co{s^4}x} \right)^2} - 2{\sin ^4}x.co{s^4}x\\ = 1 + {\sin ^4}x.co{s^4}x - 4{\sin ^2}x.co{s^2}x\\ = 1 + 2{\sin ^2}x.co{s^2}x.\left( {{{\sin }^2}x.co{s^2}x - 2} \right)\\ = 1 + \frac{{{{\sin }^2}2x}}{2}.\left( {\frac{{{{\sin }^2}2x}}{4} - 1} \right)\end{array}\)

Đặt t = sin²2x (0 < t < 1)

Phương trình đã cho có dạng:

\(\begin{array}{l}1 + \frac{t}{2}\left( {\frac{t}{4} - 2} \right) = \frac{1}{8}\\ \Leftrightarrow 8 + {t^2} - 8t = 1\\ \Leftrightarrow {t^2} - 8t + 7 = 0\\ \Rightarrow \left( {t - 1} \right)\left( {t - 7} \right) = 0\\ \Leftrightarrow t = 1\,\,(TM)\end{array}\)

Với t = 1 ta có:

\({\sin ^2}2x = 1 \Rightarrow \left\{ \begin{array}{l}x = \frac{\pi }{4} + k\pi \\x = - \frac{\pi }{4} + k\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\).