ta có sinx-cosx =[tex]\frac{1-\sqrt{3}}{2}[/tex]
<=> [tex]\sqrt{2}sin(x-\frac{\pi }{4})=\frac{1-\sqrt{3}}{2}[/tex]
<=> [tex]sin(x-\frac{\pi }{4})=\frac{-\sqrt{6}+\sqrt{2}}{4}[/tex]
<=> [tex]x-\frac{\pi }{4}=- \frac{\pi }{12}+ k2\pi[/tex] hoặc [tex]x-\frac{\pi }{4}=pi + \frac{\pi }{12}+ k2\pi[/tex]
=> x=[tex]\frac{\pi }{3}+k2\pi[/tex] và x = [tex]\frac{4\pi }{3}+k2\pi[/tex]