Solving the trigonometric equation $\sqrt{2}\sin(⁡2x)=-\sqrt{3\sin(⁡x) + 3\cos(⁡x) + 8\cos^4⁡(x-\pi/4)}$

Using auxiliary angle gives $$ \begin{aligned} \sqrt{2} \sin (2 x) &=-\sqrt{3 \sin x+3 \cos x+8 \cos ^{4}\left(x-\frac{\pi}{4}\right)} \\ &=-\sqrt{3 \sqrt{2} \cos \left(x-\frac{\pi}{4}\right)+8 \cos ^{4}\left(x-\frac{\pi}{4}\right)} \end{aligned} $$ Let $ y=x-\dfrac{\pi}{4}$, then $$\sqrt{2} \cos \left(2 y\right)=-\sqrt{3 \sqrt{2} \cos y+8 \cos ^{4} y}$$

Squaring both sides yields $$ 2 \cos ^{2}(2 y)=3 \sqrt{2} \cos y+8 \cos ^{4} y $$ Using double-angle formula gives

$$ \begin{array}{l} 2\left(2 \cos ^{2} y-1\right)^{2}=3 \sqrt{2} \cos y+8 \cos ^{4} y \\ 8 \cos ^{4} y-8 \cos ^{2} y+2=3 \sqrt{2} \cos y+8 \cos ^{4} y \\ 8 \cos ^{2} y+3 \sqrt{2} \cos y-2=0 \\ \displaystyle \cos y=\frac{-3 \sqrt{2} \pm \sqrt{82}}{16} \end{array} $$ After checking for $\sin(2x)<0$, we can conclude that the solutions are$$ \begin{array}{l} \\ \displaystyle x=\frac{(8 n+1) \pi}{4} \pm \cos ^{-1}\left(\frac{\sqrt{82}-3 \sqrt{2}}{16}\right), \end{array} $$ where $ n\in Z$.