Rozwiązanie:
[tex]1-sin2x=cosx-sinx\\1-2sinxcosx-cosx+sinx=0\\sin^{2}x-2sinxcosx+cos^{2}x+sinx-cosx=0\\(sinx-cosx)^{2}+(sinx-cosx)=0\\(sinx-cosx)(sinx-cosx+1)=0\\sinx-cosx=0 \vee sinx-cosx=-1\\sinx-sin(\frac{\pi}{2}-x) =0 \vee sinx-sin(\frac{\pi}{2} -x)=-1\\2cos\frac{x+\frac{\pi}{2}-x }{2} *sin\frac{x+x-\frac{\pi}{2} }{2} =0 \vee 2cos\frac{x+\frac{\pi}{2}-x }{2} *sin\frac{x+x-\frac{\pi}{2} }{2}=-1\\2cos\frac{\pi}{4} *sin(x-\frac{\pi}{4})=0 \vee 2cos\frac{\pi}{4} *sin(x-\frac{\pi}{4})=-1\\[/tex]
[tex]\sqrt{2} sin(x-\frac{\pi}{4}) =0 \vee \sqrt{2} sin(x-\frac{\pi}{4})=-1\\sin(x-\frac{\pi}{4}) =0 \vee sin(x-\frac{\pi}{4} )=-\frac{\sqrt{2} }{2} \\x-\frac{\pi}{4} =k\pi \vee x-\frac{\pi}{4}=-\frac{\pi}{4} +2k\pi \vee x-\frac{\pi}{4}=-\frac{3\pi }{4} +2k\pi \\x=\frac{\pi}{4}+k\pi \vee x=2k\pi \vee x=-\frac{\pi}{2} +2k\pi[/tex]
Uwzględniając przedział dostaniemy:
[tex]x \in [ \ 0,\frac{\pi}{4} , \frac{5\pi}{4} ,\frac{3\pi}{2} ,2\pi ][/tex]