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 $0<x≤2π$)
 A: $\sin(x)$ is an odd function, so $\sin(x) = -\sin(-x)$.  This means that
\begin{align*}
\sin\left(x-\frac{\pi}{4}\right)=\frac{1+\sqrt3}{2\sqrt2} &\iff -\sin\left(-x+\frac{\pi}{4}\right)=\frac{1+\sqrt3}{2\sqrt2}\\
&\iff\sin\left(-x+\frac{\pi}{4}\right)=-\frac{1+\sqrt3}{2\sqrt2}.
\end{align*}
In other words, both equations are equivalent.  I'm not sure why WolframAlpha makes this transformation first, it's probably related to whatever algorithm it is using to solve the equation.  For solving on paper it doesn't matter which approach you take.
To solve the equation in its original form, we first note that the equation is in the form $\sin(\theta) = a > 0$, which means that $\theta$ is in the first or second quadrant.  Since the domain of $\arcsin(x)$ is the first and fourth quadrants, we will need to consider the solutions
$$\theta = \arcsin(a)\quad \text{and}\quad \theta = \pi - \arcsin(a).$$
With that in mind, using the first equation, we have
$$
\sin\left(x-\frac{\pi}{4}\right)=\frac{1+\sqrt3}{2\sqrt2} \implies x-\frac{\pi}{4} =\begin{cases}  \arcsin\left(\frac{1+\sqrt3}{2\sqrt2}\right)\\
\pi - \arcsin\left(\frac{1+\sqrt3}{2\sqrt2}\right).
\end{cases}
$$
Solving for $x$ gives us
$$x= \begin{cases}\arcsin\left(\frac{1+\sqrt3}{2\sqrt2}\right) + \frac{\pi}{4} &= \frac{2\pi}{3}\\\pi - \arcsin\left(\frac{1+\sqrt3}{2\sqrt2}\right) + \frac{\pi}{4} &= \frac{5\pi}{6}.\end{cases}$$
A: 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}
