Giải phương trình sau: 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

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

⇔ $\left[\begin{array}{l}x=k\pi \left(k\in \mathbb{Z}\right)\\ \sin x\left({\sin }^{2}x+{\cos }^{2}x\right)-\cos x=0\end{array}\right.$

⇔ $\left[\begin{array}{l}x=k\pi \left(k\in \mathbb{Z}\right)\\ \sin x-\cos x=0\end{array}\right.$

⇔ $\left[\begin{array}{l}x=k\pi \left(k\in \mathbb{Z}\right)\\ \sin \left(x-\frac{\pi }{4}\right)=0\end{array}\right.$

⇔ $\left[\begin{array}{l}x=k\pi \left(k\in \mathbb{Z}\right)\\ \sin \left(x-\frac{\pi }{4}\right)=0\end{array}\right.$

⇔ $\left[\begin{array}{l}x=k\pi \\ x-\frac{\pi }{4}=k\pi \end{array}\right.\left(k\in \mathbb{Z}\right)$

⇔ $\left[\begin{array}{l}x=k\pi \\ x=\frac{\pi }{4}+k\pi \end{array}\right.\left(k\in \mathbb{Z}\right)$ .