Trigonometric inequality $\cos x+ \sin x>0$ Solve the inequality: $\cos x+ \sin x >0$ 
Why can't I square this to get $\sin 2x>0$? And what is the first step here then?

$$\cos x+ \sin x >0\Rightarrow \frac{\cos}{\sqrt2}+\frac{\sin x}{\sqrt2}=\sin(\frac{\pi}{4}+x)\gt0$$
It follows $$x\in\bigcup\space\left]-\frac{\pi}{4}+2k\pi,\frac{3\pi}{4}+2k\pi\space\right[$$
