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 ?

The inequality
$$
\sin \left( x - \frac{\pi}{3} \right) > \frac{\sqrt{2}}{2}
$$
implies that
$$
\frac{\pi}{4} < x - \frac{\pi}{3} < \frac{3\pi}{4}.
$$
Adding $\frac{\pi}{3}$ to all three expressions yields
$$
\frac{7\pi}{12} < x < \frac{13\pi}{12}.
$$
If you impose the initial restriction, then the upper bound is $\pi$.
