Giải phương trình: sin⁴x + \({\cos ^4}\left( {x + \frac{\pi }{4}} \right) = \frac{1}{4}\).

sin⁴x + \({\cos ^4}\left( {x + \frac{\pi }{4}} \right) = \frac{1}{4}\)

⇔ sin⁴x + \({\left( {\cos x.\frac{1}{{\sqrt 2 }} - \sin x.\frac{1}{{\sqrt 2 }}} \right)^4} = \frac{1}{4}\)

⇔ sin⁴x + \(\frac{1}{4}{\left( {\cos x - \sin x} \right)^4} = \frac{1}{4}\)

⇔ sin⁴x + \(\frac{1}{4}{\left[ {{{\left( {\cos x - \sin x} \right)}^2}} \right]^2} = \frac{1}{4}\)

⇔ sin⁴x + \(\frac{1}{4}{\left( {{{\cos }^2}x - 2\sin x\cos x + {{\sin }^2}x} \right)^2} = \frac{1}{4}\)

⇔ sin⁴x + \(\frac{1}{4}{\left( {1 - 2\sin x\cos x} \right)^2} = \frac{1}{4}\)

⇔ sin⁴x + sin²xcos²x – sinxcosx = 0

⇔ sinx(sin³x + sinxcos²x – cosx) = 0

⇔ \(\left[ \begin{array}{l}\sin x = 0\left( 1 \right)\\{\sin ^3}x + \sin x{\cos ^2}x - \cos x = 0\left( 2 \right)\end{array} \right.\)

(1): x = kπ

(2): sinx(sin² + cos²x) – cosx = 0

⇔ sinx – cosx = 0

⇔ \(\sin \left( {x - \frac{\pi }{4}} \right) = 0\)

⇔ \(x = \frac{\pi }{4} + k\pi \left( {k \in \mathbb{Z}} \right)\)

Vậy phương trình có nghiệm: \(x = \frac{\pi }{4} + k\pi \left( {k \in \mathbb{Z}} \right)\).