Giải phương trình: \({\sin ^4}x + {\sin ^4}\left(...

B

- Hướng dẫn giải

Lời giải chi tiết:

\(\begin{array}{l}{\sin ^4}x + {\sin ^4}\left( {x + \dfrac{\pi }{4}} \right) + {\cos ^4}\left( {x + \dfrac{\pi }{4}} \right) = \dfrac{9}{8}\\ \Leftrightarrow {\sin ^4}x + {\left[ {\dfrac{1}{{\sqrt 2 }}\left( {\sin \,x + \cos x} \right)} \right]^4} + {\left[ { - \dfrac{1}{{\sqrt 2 }}\left( {\sin \,x - \cos x} \right)} \right]^4} = \dfrac{9}{8}\\ \Leftrightarrow 4{\sin ^4}x + {\left( {\sin \,x + \cos x} \right)^4} + {\left( {\sin \,x - \cos x} \right)^4} = \dfrac{9}{2}\\ \Leftrightarrow 4{\sin ^4}x + {\left[ {{{\left( {\sin \,x + \cos x} \right)}^2}} \right]^2} + {\left[ {{{\left( {\sin \,x - \cos x} \right)}^2}} \right]^2} = \dfrac{9}{2}\\ \Leftrightarrow 4{\sin ^4}x + {\left( {1 + \sin 2x} \right)^2} + {\left( {1 - \sin 2x} \right)^2} = \dfrac{9}{2}\\ \Leftrightarrow 4{\sin ^4}x + 2{\sin ^2}2x - \dfrac{5}{2} = 0\\ \Leftrightarrow 4{\sin ^4}x + 8{\sin ^2}x.{\cos ^2}x - \dfrac{5}{2} = 0\end{array}\)

Đặt \({\sin ^2}x = t\,\,\,\,\left( {\left| t \right| \le 1} \right)\)

\(\begin{array}{l}4{t^2} - 8t\left( {1 - t} \right) - \dfrac{5}{2} = 0 \Leftrightarrow  - 4{t^2} + 8t - \dfrac{5}{2} = 0 \Leftrightarrow \left[ \begin{array}{l}t = \dfrac{{4 + \sqrt 6 }}{4}\,\,\,\left( {ktm} \right)\\t = \dfrac{{4 - \sqrt 6 }}{4}\,\,\,\left( {tm} \right)\end{array} \right.\\ \Leftrightarrow {\sin ^2}x = \dfrac{{4 - \sqrt 6 }}{4} \Leftrightarrow \dfrac{1}{2}\left( {1 - \cos 2x} \right) = \dfrac{{4 - \sqrt 6 }}{4}\\ \Leftrightarrow \cos 2x = \dfrac{{ - 2 + \sqrt 6 }}{2} \Leftrightarrow x =  \pm \dfrac{1}{2}\arccos \dfrac{{ - 2 + \sqrt 6 }}{2} + k\pi \,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)