Question

Find the range of y such that the equation in x, $y+\cos x=\sin x$ has a real solution. For $y=1$, find x such that $0<x<2\pi$.

Answer 1

$y=\sin x-\cos x$
or $\frac{y}{\sqrt{2}}=\frac{1}{\sqrt{2}}\sin x-\frac{1}{\sqrt{2}}\cos x$
or $\frac{y}{\sqrt{2}}=\sin (x-\frac{\pi }{4})$. For real solution
$-1\le (x-\frac{\pi }{4})\le 1$ or $-1\le \frac{y}{\sqrt{2}}\le 1$
If $y=1$, then $\sin x-\cos x=1$
or $\frac{1}{\sqrt{2}}=\sin (x-\frac{\pi }{4})=\sin \frac{\pi }{4}$
$\therefore x-\frac{\pi }{4}=n\pi +(-1{)}^n\frac{\pi }{4}$
n even, $x=2r\pi +\frac{\pi }{2}$; n odd, $x=(2r+1)\pi$
$\therefore x=\pi /2,\pi$ as $0<x<2\pi$.