Question

The number of value of $x$ in $[0,2\pi ]$ satisfying the equation $|\cos x-\sin x|\ge \sqrt{2}$, is

Answer 1

Given equation is $|\cos x-\sin x|\ge \sqrt{2}$
Since $|\cos x-\sin x|\le \sqrt{1+1}=\sqrt{2}$,
$\therefore$ we must have $|\cos x-\sin x|=\sqrt{2}$
$\Rightarrow \left|\cos (x+\frac{\pi }{4})\right|=1\Rightarrow \cos (x+\frac{\pi }{4})=-1,1$
$\therefore x+\frac{\pi }{4}=0,2\pi ,4\pi ,6\pi ....\pi ,3\pi ,5\pi ,...$
$\therefore x=\frac{3\pi }{4},\frac{7\pi }{4}$