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 ?
 A: 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 $\frac{7\pi}{12} < x < \pi$.
A: 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 $\pi$.
A: We need $$2\sin\left(x-60^\circ\right)-\sqrt2>0$$
But as $\sin45^\circ=\frac1{\sqrt2},$ it essentially implies and is implied by $$2\sin\left(x-60^\circ\right)-2\sin45^\circ>0$$  
using Prosthaphaeresis Formulas,
$$\sin\left(x-60^\circ\right)-\sin45^\circ=2\sin\dfrac{x-105^\circ}2\cos\frac{x-15^\circ}2\ \ \ \ (1)$$
Now, $$\sin\frac{x-105^\circ}2>0\iff n360^\circ<\frac{x-105^\circ}2<n360^\circ+180^\circ$$
$$\iff n720^\circ+105^\circ<x<n720^\circ+(540+105)^\circ$$
Setting $n=0, 105^\circ<x<(540+105)^\circ$
and setting $n=-1,$
$(-720+105)^\circ<x<(-720+540+105)^\circ\iff -615^\circ<x<-75^\circ$
As we have $-180^\circ<x<180^\circ,$
$\displaystyle\sin\frac{x-105^\circ}2>0\iff-180^\circ<x<-75^\circ$ or $105^\circ<x<180^\circ\  \ \ \ (2)$
Again, $$\cos\frac{x-15^\circ}2>0\iff m360^\circ-90^\circ<\frac{x-15^\circ}2<m360^\circ+90^\circ$$
$$m720^\circ-165^\circ<x<m720^\circ+195^\circ$$
Set $n=0$ as $-180^\circ<x<180^\circ,$
$\displaystyle \cos\frac{x-15^\circ}2>0\iff -165^\circ<x<180^\circ \  \ \ \ (3)$
So, $(1)$ will be $>0$ 
if $105^\circ<x<180^\circ$(where  both the multipliers in $(2)<(3)$  are positive ) 
or if $-180^\circ<x<-165^\circ$ (where both the multipliers are negative ) 
