<s>[INST] Why do we multiply by $-1$ here? WolframAlpha solves $$\sin\left(x-\fracπ4\right)=\frac{1+\sqrt3}{2\sqrt2}$$ by multiplying by $-1$ as such:
$$\sin\left(-x+\fracπ4\right)=-\frac{1+\sqrt3}{2\sqrt2}$$ then arcsin both sides, etc. but why multiplying by $-1$ and not just directly find the answer? Mainly say this because I've tried without multiplying by $-1$ and haven't managed to got $\frac56 π$ which is the answer (in $0<x≤2π$)
 [/INST] Apply  trig addition identity,we can do this step by step as the following (there is no requirement to multiply -1 for both sides):
\begin{align}\sin\left(x-\fracπ4\right)=\frac{1+\sqrt3}{2\sqrt2} \end{align}
\begin{align}  \sin x\cos\frac{\pi}{4}-\cos x\sin \frac{π}{4}=\frac{1+\sqrt3}{2\sqrt2}  \end{align}\begin{align}(\sin x-\cos x)\frac{\sqrt 2}{2}=\frac{1+\sqrt3}{2\sqrt2}\end{align} \begin{align}\sin x-\cos x=\frac{1}{2}+\frac{\sqrt3}{2} \end{align}
\begin{align}x = \frac{2\pi}{3},\frac{5\pi}{6}  ( x\in (0,2\pi])  \end{align}
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