Doubts in Trigonometrical Inequalities I'm now studying Trigonometrical Inequalities, and I've just got struck when I have modified arguments to my trigonometrical functions, for example:
$\sqrt{2} - 2\sin\left(x - \dfrac{\pi}{3} \right) < 0$ when $-\pi < x < \pi$
With some work I've got: $\sin\left(x - \dfrac{\pi}{3} \right) > \dfrac{\sqrt{2}}{2}$
To find bounds: $\sin(x) = \dfrac{\sqrt{2}}{2},\ x = \dfrac{\pi}{4},\ \dfrac{3\pi}{4}$
Resolving to $x + \dfrac{\pi}{3} \implies \dfrac{\pi}{4} < x < \dfrac{3\pi}{4}$
But wolfram gives a way different result, where's my mistake ?

First of all, you said that $\sin x = \frac{\sqrt{2}}{2}$ when $x = \frac{\pi}{4}, \frac{3\pi}{4}$ then concluded that $\sin x > \frac{\sqrt{2}}{2}$ when $\frac{\pi}{4} < x < \frac{3\pi}{4}$. While this is true, you should give some explanation here as it could be the case that $\sin x < \frac{\sqrt{2}}{2}$ for $\frac{\pi}{4} < x < \frac{3\pi}{4}$.
As $\sin x > \frac{\sqrt{2}}{2}$ for $\frac{\pi}{4} < x < \frac{3\pi}{4}$, $\sin(x - \frac{\pi}{3}) > \frac{\sqrt{2}}{2}$ for $\frac{\pi}{4} < x - \frac{\pi}{3} < \frac{3\pi}{4}$. By adding $\frac{\pi}{3}$ to each term in the inequality, we have $\frac{7\pi}{12} < x < \frac{13\pi}{12}$. 
So, for every $k \in \mathbb{Z}$, we have $\sqrt{2} - 2\sin(x-\frac{\pi}{3}) > 0$ for $\frac{7\pi}{12}+2k\pi < x < \frac{13\pi}{12}+2k\pi$. 
For $k = -1$ we have $-\frac{17\pi}{12} < x < -\frac{11\pi}{12}$, and for $k = 0$ we have $\frac{7\pi}{12} < x < \frac{13\pi}{12}$. As we are looking for $x$ which satisfy $-\pi < x < \pi$, these are the only $x$ we need to consider (for any other $k$, the corresponding inequalities do not allow for $x$ which also satisfy $-\pi < x < \pi$). 
If $x$ satisfies $-\frac{17\pi}{12} < x < -\frac{11\pi}{12}$ and $-\pi < x < \pi$, then $-\pi < x < -\frac{11\pi}{12}$.
If $x$ satisfies $\frac{7\pi}{12} < x < \frac{13\pi}{12}$ and $-\pi < x < \pi$, then $\frac{7\pi}{12} < x < \pi$. 
Therefore, for $-\pi < x < \pi$, $\sqrt{2} -2\sin(x-\frac{\pi}{3}) > 0$ for $-\pi < x < -\frac{11\pi}{12}$ and $\frac{7\pi}{12} < x < \pi$.
