I'm solving the following trigonometric inequation ( $cos(x)-sin(x)+1 \gt 0$) on the interval $[0,2\pi]$. And I found that

$-\pi +2k \pi \lt x \lt {\pi\over 2} + 2k\pi $, $k\in \mathbb Z$

So, if I'm not wrong I have to find $k$ such that the following inequation holds:

$ 0 \le -\pi +2k \pi \lt {\pi\over 2} + 2k\pi \le 2\pi $, $k\in \mathbb Z$

But I'm not able to solve this inequation, can someone explain me how to solve it or maybe explain me with a more general inequation.

I know I can just try several value for $k$ like $k=0$ and $k=1$ but I dont know how to deal such inequation with bigger value.


  • $\begingroup$ Hint : $\sin x - \cos x = \sqrt{2}\sin(x - \pi/4)$ $\endgroup$ – S.H.W May 22 '18 at 18:01
  • $\begingroup$ I used this To solve the inequation and find what $x$ must satisfies but how can I find algebraiclly the $k$ ? $\endgroup$ – Sami Mir May 22 '18 at 18:35

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