<s>[INST] Solving the trigonometric equation $\sqrt{2}\sin(⁡2x)=-\sqrt{3\sin(⁡x) + 3\cos(⁡x) + 8\cos^4⁡(x-\pi/4)}$ My problem is to solve the following equation:
$$\sqrt{2}\sin(⁡2x)=-\sqrt{3\sin(⁡x) + 3\cos(⁡x) + 8\cos^4⁡(x-\pi/4)}$$
I've narrowed it down to
$$3\sin(⁡x) + 3\cos(⁡x) + 2 + 8\sin(⁡x)\cos(⁡x) = 0.$$
That's as far as I can get. It seems so simple, but I just can't find the next step(s).
 [/INST] 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$.
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